AE circulation of Ethereum in Europe
Publish: 2021-05-09 06:16:25
1. That is, MLM (without any physical goods and investment procts in essence)
direct selling, secondary, tertiary or provincial or municipal level, through temptation, to attract people to join MLM refers to: popular point refers to the behavior of selling or promoting goods as the name, county-level and other regional agents, and collecting membership fees as the main way of profit: the sales system built by groups or enterprises, Indivial or team development at the national level
direct selling, secondary, tertiary or provincial or municipal level, through temptation, to attract people to join MLM refers to: popular point refers to the behavior of selling or promoting goods as the name, county-level and other regional agents, and collecting membership fees as the main way of profit: the sales system built by groups or enterprises, Indivial or team development at the national level
2. Example 1 is known: as shown in Figure 1-1, in the quadrilateral ABCD, BC > Ba, ad = CD, BD bisects ∠ ABC,
verification: ∠ a + C = 180 °
analysis: because the horizontal angle is equal to 180 °， Therefore, we should consider transforming two angles which are not together into equal angles by congruence, and there is no congruent triangle in the graph. Therefore, the key to solve the problem is to construct isosceles triangle, which can be realized by "truncation and complement method". "
prove: intercept be = AB on BC, connect De, and then take the midpoint m of EC, Connecting DM
∵ AB = be
and ∵ BD bisecting ∠ ABC a d
∵ abd = ∠ EBD
in △ abd and △ EBD,
AB = be
∠ abd = ∠ EBD
BD = BD B e M C
{abd ≌ △ EBD (SAS) is shown in Figure 1-1
} ad = ed ∠ a = ∠ bed,
∵AD = DC ， ∴ED = DC ∴∠ C = ∠DEC
∴∠A + ∠C = ∠BED +∠DEC = 180 °< As shown in Figure 2-2, AE / / BC, ad and BD share & eth equally; EAB、Ð CBA, EC, point D
verification: ab = AE + BC
analysis 1: to prove that ab = AE + BC, observe that AD and BD are bisectors, so the Daed can be folded along a, so it is necessary to add an auxiliary line to intercept BF = BC on AB, and only need to dece AF = AE, then the problem can be solved, so how to dece AF = AE becomes the key to solve the problem. Because the sum of internal angles of Daed, DADB and DBD is 180 °， And & eth; EDC=180 °， Because AE / / be, therefore & eth; E+Ð C=180 ° Thus & eth; EAB+Ð CBA=180 °， Ad and BD are bisectors of angles, and we can dece & eth; 1+Ð 4=90 °， Thus, we can dece & eth; ADB=90 °， Thus & eth; 6+Ð 8=90 ° If we can dece & eth; 7=Ð 8, then we only need to dece Daed ≌ dafd, thus we can dece AE = AF, BC = BF, & eth; 1=Ð 2. BD is a common edge, so we can dece dbfd ≌ DBCD, then & eth; 5=Ð 6, e to & eth; 5+Ð 7=90 ° So & eth; 6+Ð 7=90 °， Due to & eth; 6+Ð 8=90 °， Thus & eth; 7=Ð Therefore, ad can be the common edge, & eth; 3=Ð 4 dece Daed ≌ dafd, thus the idea is smooth, dece AE = AF, and dece AB = AE + BC by equivalent substitution
proof 1: intercept BF = BC on AB and link DF
BD is & eth; Bisector of ABC & eth; 1=Ð 2
in dbdf and DBDC,
(common side)
{dbdf ≌ DBDC (SAS) is shown in Figure 2-2
} & eth; 5=Ð 6 (the corresponding angles of congruent triangles are equal)
! & eth; 3+Ð 8+Ð E=Ð 4+Ð 1+Ð 5+Ð 7=Ð 2+Ð 6+Ð C=180 ° The sum theorem of internal angles of triangle)
{eth; E+Ð EAB+Ð ABC+Ð C+Ð EDC=540 °
and ﹣ AE / / BC ﹣ eth; E+Ð C=180 ° Two lines are parallel and complementary)
and ∵ & eth; EDC=180 ° ∴Ð 1+Ð 2+Ð 3+ Ð 4=180 °
ad is & eth; The bisector of EAB & eth; 3=Ð 4
∴Ð 1+Ð 4=90 ° ∴Ð 5+Ð 7=90 ° The sum theorem of internal angles of triangle)
{eth; 6+Ð 8=90 ° ∵Ð 5=Ð 6 ∴ Ð 7=Ð 8
in Daed and dafd,
{Daed ≌ dafd (ASA)
} AE = AF (the corresponding sides of congruent triangles are equal)
≌ AF + FB = AB
} AE = FB = AE + BC = AB
that is, ab = AE + BC
analysis 2: extend the extension line of BC intersecting ad to F. To prove AB = AE + BC, we only need to prove BF = AB and dece CF = AE. In order to prove CF = AE, we only need to dece two triangles Daed ≌ dfcd with CF and AE e to & eth; 5 = 6, AE / / BC, so & eth; 3=Ð F. If we want to dece ad = FD, it becomes the key to solve the problem °，& ETH; E+Ð BCE=180 °， Therefore, it can be known that & eth; EAB+Ð CBA=180 °， Ad and BD are & eth; EAB、Ð CBA's bisector, thus can be introced & eth; 1+Ð 4=90 °， Therefore & eth; ADB=90 °， Then & eth; EDB=90 °， At this point, they can dece dabd ≌ dfbd according to ASA by observing the figure, and thus dece ad = FD, forming the idea
proof 2: as shown in Fig. 2-3, the sum of the internal angles of the three triangles in Daed, DADB and DBDC is 540 ° The sum theorem of internal angles of triangles)
and ∵ & eth; EDC=180 ° Horizontal angle definition & eth; E+Ð C+Ð EAB+Ð ABC=180 °
AE / / BC (two lines are parallel and complementary)
■ & eth; 3+Ð 4+Ð 1+Ð 2=180 °
AD and BD are & eth respectively; EAB、Ð The bisector of ABC
& eth; 3=Ð 4，Ð 1=Ð 2 (angular bisector definition)
﹣ eth; 1+Ð 4=90 ° ∴Ð ADB=90 ° The sum theorem of internal angles of triangle)
{eth; BDF=90 ° In DADB and dbdf,
≌ DADB ≌ dbdf (ASA)
≌ ad = FD, ab = FB, & eth; 4=Ð F (opposite sides of congruent triangle, corresponding angles are equal) as shown in Fig. 2-3
in Daed and dfcd, it is known that ad is the angular bisector of △ ABC, AB > AC, as shown in Fig. 3-1,
verification: AB-AC > bd-dc
analysis: in order to prove AB-AC > bd-dc, we need to transform the difference between AB and AC, the difference between BD and DC or their equal quantity into the edge of the same triangle, and then use the relationship between the three sides of the triangle to prove it
proof: Method 1: intercept AE = AC on AB and connect ed. A
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ ade and △ ADC, E
AE = AC
ead = ∠ DAC B D C
ad = ad as shown in Figure 3-1
{△ ade ≌ △ ADC (SAS)} de = D C
in △ abd, Be > bd-de (the difference between the two sides of the triangle is less than the third side)
that is, ab-ae > bd-dc
Ψ AB-AC > bd-dc (equivalent substitution)
method 2: short complement method
extend AC to point E, so that AE = ab, Connecting de a
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ bad and △ ead,
AB = AEC
∠ bad = ∠ DAC B D E
ad = ad
{ade ≌ delta ADC (SAS)} D B = D E, as shown in figure 3-2
in △ abd, EC > de-dc (the difference between the two sides of the triangle is less than the third side)
that is ae-ac > de-dc AB-AC > bd-dc
example 4 shows that, as shown in Figure 4-1, in △ ABC, ∠ C = 2 ∠ B, ∠ 1 = 2.
verification: ab = AC + CD.
analysis: from the conclusion analysis, "truncation" or "complement" can realize the transformation of the problem, that is, extending AC to e makes CE = CD, It is proved that: Method 1 (short complement method)
prolongs AC to e, so that DC = CE, then ∠ CDE = ∠ CED, as shown in Figure 4-2
Figure 4-2
achb = 2 ∠ e,
∫ ACB = 2 ∠ B, ∫ B = e,
in △ abd and △ AED,
{abd ≌ △ AED (AAS)
, ≌, And the AE = AC + CE = AC + DC, < br /
< br /
< br /
< br / < br / < br / < br /
< br /
< br / < br / < br / < method 2 (truncation method) < br / < br /
in the Figure 4-3, as shown in Figure 4-3 < br / < br > < br / < br /
in △ AFD and △ ACD, < br /
< br /
< br / < br / < br > < br /
< br / < br / < br > < br / < br / < br > < br / < br / < br > < br / < br > < br / < br / < br > < br / < br / < br / < in △ AFD 87808780 8780 87808780b,
∴FD=FB. ∵AB=AF+FB=AC+FD，
∴AB=AC+CD.
verification: ∠ a + C = 180 °
analysis: because the horizontal angle is equal to 180 °， Therefore, we should consider transforming two angles which are not together into equal angles by congruence, and there is no congruent triangle in the graph. Therefore, the key to solve the problem is to construct isosceles triangle, which can be realized by "truncation and complement method". "
prove: intercept be = AB on BC, connect De, and then take the midpoint m of EC, Connecting DM
∵ AB = be
and ∵ BD bisecting ∠ ABC a d
∵ abd = ∠ EBD
in △ abd and △ EBD,
AB = be
∠ abd = ∠ EBD
BD = BD B e M C
{abd ≌ △ EBD (SAS) is shown in Figure 1-1
} ad = ed ∠ a = ∠ bed,
∵AD = DC ， ∴ED = DC ∴∠ C = ∠DEC
∴∠A + ∠C = ∠BED +∠DEC = 180 °< As shown in Figure 2-2, AE / / BC, ad and BD share & eth equally; EAB、Ð CBA, EC, point D
verification: ab = AE + BC
analysis 1: to prove that ab = AE + BC, observe that AD and BD are bisectors, so the Daed can be folded along a, so it is necessary to add an auxiliary line to intercept BF = BC on AB, and only need to dece AF = AE, then the problem can be solved, so how to dece AF = AE becomes the key to solve the problem. Because the sum of internal angles of Daed, DADB and DBD is 180 °， And & eth; EDC=180 °， Because AE / / be, therefore & eth; E+Ð C=180 ° Thus & eth; EAB+Ð CBA=180 °， Ad and BD are bisectors of angles, and we can dece & eth; 1+Ð 4=90 °， Thus, we can dece & eth; ADB=90 °， Thus & eth; 6+Ð 8=90 ° If we can dece & eth; 7=Ð 8, then we only need to dece Daed ≌ dafd, thus we can dece AE = AF, BC = BF, & eth; 1=Ð 2. BD is a common edge, so we can dece dbfd ≌ DBCD, then & eth; 5=Ð 6, e to & eth; 5+Ð 7=90 ° So & eth; 6+Ð 7=90 °， Due to & eth; 6+Ð 8=90 °， Thus & eth; 7=Ð Therefore, ad can be the common edge, & eth; 3=Ð 4 dece Daed ≌ dafd, thus the idea is smooth, dece AE = AF, and dece AB = AE + BC by equivalent substitution
proof 1: intercept BF = BC on AB and link DF
BD is & eth; Bisector of ABC & eth; 1=Ð 2
in dbdf and DBDC,
(common side)
{dbdf ≌ DBDC (SAS) is shown in Figure 2-2
} & eth; 5=Ð 6 (the corresponding angles of congruent triangles are equal)
! & eth; 3+Ð 8+Ð E=Ð 4+Ð 1+Ð 5+Ð 7=Ð 2+Ð 6+Ð C=180 ° The sum theorem of internal angles of triangle)
{eth; E+Ð EAB+Ð ABC+Ð C+Ð EDC=540 °
and ﹣ AE / / BC ﹣ eth; E+Ð C=180 ° Two lines are parallel and complementary)
and ∵ & eth; EDC=180 ° ∴Ð 1+Ð 2+Ð 3+ Ð 4=180 °
ad is & eth; The bisector of EAB & eth; 3=Ð 4
∴Ð 1+Ð 4=90 ° ∴Ð 5+Ð 7=90 ° The sum theorem of internal angles of triangle)
{eth; 6+Ð 8=90 ° ∵Ð 5=Ð 6 ∴ Ð 7=Ð 8
in Daed and dafd,
{Daed ≌ dafd (ASA)
} AE = AF (the corresponding sides of congruent triangles are equal)
≌ AF + FB = AB
} AE = FB = AE + BC = AB
that is, ab = AE + BC
analysis 2: extend the extension line of BC intersecting ad to F. To prove AB = AE + BC, we only need to prove BF = AB and dece CF = AE. In order to prove CF = AE, we only need to dece two triangles Daed ≌ dfcd with CF and AE e to & eth; 5 = 6, AE / / BC, so & eth; 3=Ð F. If we want to dece ad = FD, it becomes the key to solve the problem °，& ETH; E+Ð BCE=180 °， Therefore, it can be known that & eth; EAB+Ð CBA=180 °， Ad and BD are & eth; EAB、Ð CBA's bisector, thus can be introced & eth; 1+Ð 4=90 °， Therefore & eth; ADB=90 °， Then & eth; EDB=90 °， At this point, they can dece dabd ≌ dfbd according to ASA by observing the figure, and thus dece ad = FD, forming the idea
proof 2: as shown in Fig. 2-3, the sum of the internal angles of the three triangles in Daed, DADB and DBDC is 540 ° The sum theorem of internal angles of triangles)
and ∵ & eth; EDC=180 ° Horizontal angle definition & eth; E+Ð C+Ð EAB+Ð ABC=180 °
AE / / BC (two lines are parallel and complementary)
■ & eth; 3+Ð 4+Ð 1+Ð 2=180 °
AD and BD are & eth respectively; EAB、Ð The bisector of ABC
& eth; 3=Ð 4，Ð 1=Ð 2 (angular bisector definition)
﹣ eth; 1+Ð 4=90 ° ∴Ð ADB=90 ° The sum theorem of internal angles of triangle)
{eth; BDF=90 ° In DADB and dbdf,
≌ DADB ≌ dbdf (ASA)
≌ ad = FD, ab = FB, & eth; 4=Ð F (opposite sides of congruent triangle, corresponding angles are equal) as shown in Fig. 2-3
in Daed and dfcd, it is known that ad is the angular bisector of △ ABC, AB > AC, as shown in Fig. 3-1,
verification: AB-AC > bd-dc
analysis: in order to prove AB-AC > bd-dc, we need to transform the difference between AB and AC, the difference between BD and DC or their equal quantity into the edge of the same triangle, and then use the relationship between the three sides of the triangle to prove it
proof: Method 1: intercept AE = AC on AB and connect ed. A
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ ade and △ ADC, E
AE = AC
ead = ∠ DAC B D C
ad = ad as shown in Figure 3-1
{△ ade ≌ △ ADC (SAS)} de = D C
in △ abd, Be > bd-de (the difference between the two sides of the triangle is less than the third side)
that is, ab-ae > bd-dc
Ψ AB-AC > bd-dc (equivalent substitution)
method 2: short complement method
extend AC to point E, so that AE = ab, Connecting de a
∫ ad bisection ∠ BAC ∫ bad = ∠ DAC
in △ bad and △ ead,
AB = AEC
∠ bad = ∠ DAC B D E
ad = ad
{ade ≌ delta ADC (SAS)} D B = D E, as shown in figure 3-2
in △ abd, EC > de-dc (the difference between the two sides of the triangle is less than the third side)
that is ae-ac > de-dc AB-AC > bd-dc
example 4 shows that, as shown in Figure 4-1, in △ ABC, ∠ C = 2 ∠ B, ∠ 1 = 2.
verification: ab = AC + CD.
analysis: from the conclusion analysis, "truncation" or "complement" can realize the transformation of the problem, that is, extending AC to e makes CE = CD, It is proved that: Method 1 (short complement method)
prolongs AC to e, so that DC = CE, then ∠ CDE = ∠ CED, as shown in Figure 4-2
Figure 4-2
achb = 2 ∠ e,
∫ ACB = 2 ∠ B, ∫ B = e,
in △ abd and △ AED,
{abd ≌ △ AED (AAS)
, ≌, And the AE = AC + CE = AC + DC, < br /
< br /
< br /
< br / < br / < br / < br /
< br /
< br / < br / < br / < method 2 (truncation method) < br / < br /
in the Figure 4-3, as shown in Figure 4-3 < br / < br > < br / < br /
in △ AFD and △ ACD, < br /
< br /
< br / < br / < br > < br /
< br / < br / < br > < br / < br / < br > < br / < br / < br > < br / < br > < br / < br / < br > < br / < br / < br / < in △ AFD 87808780 8780 87808780b,
∴FD=FB. ∵AB=AF+FB=AC+FD，
∴AB=AC+CD.
3. At first, people were more interested in his design concept, but there was no shortage of fresh objects in the world. But once people were cut leeks, they would slowly think about this thing. He has been there for so many years, and it should graally return to zero, Even Ma Yun's technology application can't catch up, and it's still popular in the world. It's a joke. The state's attitude towards him is not to fight or support him.
4. Currently, UOS is preferred. Because eth, EOS, AE have all gone up! UOS is the basic platform of blockchain which combines the advantages of eth + EOS.
5. If you add a route, as long as the data leading to the IP starting with 61 goes from eth0, the network card will be executed directly in DOS or written as text, and it will be executed directly in bat format, and the restart will be invalid
6. 1) It is not enough to replace hwaddr only, but also need to modify the / etc / udev / rules.d/70-persistent-net.rules file to exchange the MAC addresses of eth0 and eth1
for example:
# PCI device 0x14e4:0x1639 (bnx2)
subsystem = = & quot; net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" d4:ae:52:64:09:2f", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth1"< br /># PCI device 0x14e4:0x1639 (bnx2)
SUBSYSTEM==" net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" d4:ae:52:64:09:2d", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth0"
2) there is a difference between the two in principle. 1) it just replaces the order of kernel when loading network card
update the driver, then the new mole will be used
3) which is better, I think if there is no difference in the performance of network card, it is the same. You can choose the way you are used to
for example:
# PCI device 0x14e4:0x1639 (bnx2)
subsystem = = & quot; net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" d4:ae:52:64:09:2f", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth1"< br /># PCI device 0x14e4:0x1639 (bnx2)
SUBSYSTEM==" net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" d4:ae:52:64:09:2d", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth0"
2) there is a difference between the two in principle. 1) it just replaces the order of kernel when loading network card
update the driver, then the new mole will be used
3) which is better, I think if there is no difference in the performance of network card, it is the same. You can choose the way you are used to
7. Scheme 1:
1: first, modify the physical configuration file of network card. If there is no such file, create a new one or it from the existing environment and modify it as required
VI / etc / udev / rules. D / 70 persistent net. Rules
# PCI device 0x14e4:0x1692 (TG3)
subsystem = = & quot; net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" bc:30:5b:b1:cd:be", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth0"
keyword explanation:
attr {address} = = & quot; bc:30:5b:9c:ae:79" ## Check the MAC address of physical network card ifconfig
kernel = = & quot; eth*" ## Original NIC name
name = & quot; eth0" ## The current network card name 1234567
needs to configure the MAC address of the network card in this file, the original network card name and the network card name to be changed
2, modify the network configuration file
VI / etc / network / interfaces
auto eth0
Iface eth0 INET static
address 172.16.19.xx
netmask 255.255.255.0
this should be configured as a new network card name after modification
auto eth0 # the corresponding network card name should be correct 1234567
3, Restart the server
there is a certain risk in restarting the physical server. The server has not been restarted, the personal test environment is relatively old, and there have been cases where the server could not be restarted, or the server lost connection e to network configuration error when it was restarted. It is suggested that the server should be close to itself when modifying the network card, and should not be modified remotely, Prevent server from losing connection
scheme 3:
in / etc / default / grub, grub_ CMDLINE_ Add the parameter net.ifnames = 0, biosdevname = 0 in Linux, as shown in the following figure:
then execute the command line, update grub, and finally modify the / etc / network / interfaces file,
change the network card name to eth0
restart the system, and the network card name is changed successfully. Of course, the server will lose connection
read the full text
1: first, modify the physical configuration file of network card. If there is no such file, create a new one or it from the existing environment and modify it as required
VI / etc / udev / rules. D / 70 persistent net. Rules
# PCI device 0x14e4:0x1692 (TG3)
subsystem = = & quot; net", ACTION==" add", DRIVERS=="?*& quot;, ATTR{address}==" bc:30:5b:b1:cd:be", ATTR{dev_ id}==" 0x0", ATTR{type}==" 1", KERNEL==" eth*", NAME=" eth0"
keyword explanation:
attr {address} = = & quot; bc:30:5b:9c:ae:79" ## Check the MAC address of physical network card ifconfig
kernel = = & quot; eth*" ## Original NIC name
name = & quot; eth0" ## The current network card name 1234567
needs to configure the MAC address of the network card in this file, the original network card name and the network card name to be changed
2, modify the network configuration file
VI / etc / network / interfaces
auto eth0
Iface eth0 INET static
address 172.16.19.xx
netmask 255.255.255.0
this should be configured as a new network card name after modification
auto eth0 # the corresponding network card name should be correct 1234567
3, Restart the server
there is a certain risk in restarting the physical server. The server has not been restarted, the personal test environment is relatively old, and there have been cases where the server could not be restarted, or the server lost connection e to network configuration error when it was restarted. It is suggested that the server should be close to itself when modifying the network card, and should not be modified remotely, Prevent server from losing connection
scheme 3:
in / etc / default / grub, grub_ CMDLINE_ Add the parameter net.ifnames = 0, biosdevname = 0 in Linux, as shown in the following figure:
then execute the command line, update grub, and finally modify the / etc / network / interfaces file,
change the network card name to eth0
restart the system, and the network card name is changed successfully. Of course, the server will lose connection
read the full text
8. [ root@centos1 ~]#Ifconfig
- bash: ifconfig: command not found
first, habitually input echo $path (check the current path environment variable, which has the same function as DOS path command, and note that the commands in Linux system are case sensitive), The results are as follows:
[ root@centos1 ~]#Echo $path
/ usr / local / SBIN / usr / local / bin / usr / SBIN / usr / bin / root / bin
from the results shown above, the path / usr / SBIN to place the system management program already exists, which is the path to place the external command. Directly use ls to view the / usr / SBIN / directory, and you don't see ifconfig. What's the matter< br />
[ root@centos1 ~]#Ls / usr / SBIN /
I still don't give up and I can't find ifconfig with find command< br />
[ root@centos1 ~]# find / -name " ifconfig"
at this time, I have the bottom of my mind. I should replace ifconfig with a command. Check on the network, sure enough, IP command has been used instead of ifconfig command. The common parameters of IP command are listed below< The code is as follows:
IP [option] operation object {link | addr | route...} & lt/ p> & lt; p># The network interface information is shown in the IP link show, and the network interface information is shown in the network interface information < br /
35; IP link set eth0 UPI 35; opening the network card
35; IP link set eth0 down # IP link set eth0 down # closing the network card < br / # IP link set eth0 promise on # opening the mixed mode of the network card < br /
35; IP link set eth0 Up0 upupi 35; opening the mixed mode of the network card < mixed mode of the network card is opening the mixed mode of the mixed mode of the network card < br / < br /
# IP link set set set set set set the IP link set set eth0 to be set as the network interface to be described in the network card, which is the network interface interface interface and the next one of the network card, which is defined in the network card, which is defined in the network interface interface H0 MTU 1400 ﹥ set the maximum transmission unit of network card
﹥ IP addr show ﹥ display IP information of network card
﹥ IP addr add 192.168.0.1/24 dev eth0 ﹥ set eth0 IP address 192.168.0.1
﹥ IP addr del 192.168.0.1/24 dev eth0 ﹥ delete eth0 IP address & lt/ p> & lt; p># IP route list # view routing information
# IP route add 192.168.4.0/24 via 192.168.0.254 dev eth0 # set the gateway of 192.168.4.0 network segment to 192.168.0.254, Data goes through eth0 interface
# IP route add default via 192.168.0.254 dev eth0 # setting default gateway to 192.168.0.254
# IP route del 192.168.4.0/24 # deleting gateway of 192.168.4.0 network segment
# IP route del default # deleting default route
after entering IP addr command, it is found that enp2s0 network card (this enp2s0 is my network card) has no IP address< br />
[ root@centos1 ~]#IP addr
since there is no IP address, go directly to the / etc / sysconfig / network scripts directory to see the configuration file name of the IP information of the network card< br />
[ root@centos1 ~]# ls /etc/sysconfig/network-scripts/
ifcfg-enp2s0 ifdown-eth ifdown-post ifdown-Team ifup-aliases ifup-ipv6 ifup-post ifup-Team init.ipv6-global
ifcfg-lo ifdown-ippp ifdown-ppp ifdown-TeamPort ifup-bnep ifup-isdn ifup-ppp ifup-TeamPort network-functions
ifdown ifdown-ipv6 ifdown-routes ifdown-tunn el ifup-eth ifup-plip ifup-routes ifup-tunnel network-functions-ipv6
ifdown-bnep ifdown-isdn ifdown-sit ifup ifup-ippp ifup-plusb ifup-sit ifup-wireless
[ root@centos1 ~]#
from the results, the name of the configuration file used to save the IP information of the network card has changed from ifcfg-eth0 to ifcfg-enp2s0. Well, since you gave him such a name, I'll use it first. Cat ifcfg-enp2s0
first[ root@centos1 ~]#Cat / etc / sysconfig / network scripts / ifcfg-enp2s0
code
the code is as follows:
hwaddr = 00: e0:69:01:6a:96
type = Ethernet
bootproto = DHCP
defraute = yes
peerdns = yes
peeroutes = yes
IPv4_ FAILURE_ FATAL=no
IPV6INIT=yes
IPV6_ AUTOCONF=yes
IPV6_ DEFROUTE=yes
IPV6_ PEERDNS=yes
IPV6_ PEERROUTES=yes
IPV6_ FAILURE_ Fat = no
name = enp2s0
UUID = 5b0a7d76-1602-4e19-aee6-29f57618ca01
onboot = no
from the above configuration, we can see that although bootproto = DHCP, onboot = No. here, use VI to change onboot = no to onboot = yes, and then restart CentOS< br />
[ root@centos1 ~]#Shutdown - R
after the restart, enter the account number and password to enter the command prompt operator and continue to view the network card information with IP addr. The results are as follows:
[ root@centos1 ~]# ip add
1: lo: < LOOPBACK,UP,LOWER_ UP> mtu 65536 qdisc noqueue state UNKNOWN
link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
inet 127.0.0.1/8 scope host lo
valid_ lft forever preferred_ lft forever
inet6 ::1/128 scope host
valid_ lft forever preferred_ lft forever
2: enp2s0: < BROADCAST,MULTICAST,UP,LOWER_ UP> mtu 1500 qdisc pfifo_ fast state UP qlen 1000
link/ether 00:e0:69:01:6a:96 brd ff:ff:ff:ff:ff:ff
inet 172.8.1.200/24 brd 172.8.1.255 scope global enp2s0
valid_ lft forever preferred_ lft forever
inet6 fe80::2e0:69ff:fe01:6a96/64 scope link
valid_ lft forever preferred_ lft forever
[ root@centos1 ~]#
from the above results, we can see that the IP address assigned through DHCP is 172.8.1.200. Although it is a testing machine, we should configure a fixed IP for this machine for the convenience of remote connection in the future
open ifcfg-enp2s0 with VI, input the following parameters, and then annotate bootproto = DHCP with #
code
the code is as follows:
ipaddr0 = 172.8.1.211
prefix 0 = 24
gateway 0 = 172.8.1.1
dns1 = 172.8.1.1
the complete parameters are as follows, OK, the network is configured.
- bash: ifconfig: command not found
first, habitually input echo $path (check the current path environment variable, which has the same function as DOS path command, and note that the commands in Linux system are case sensitive), The results are as follows:
[ root@centos1 ~]#Echo $path
/ usr / local / SBIN / usr / local / bin / usr / SBIN / usr / bin / root / bin
from the results shown above, the path / usr / SBIN to place the system management program already exists, which is the path to place the external command. Directly use ls to view the / usr / SBIN / directory, and you don't see ifconfig. What's the matter< br />
[ root@centos1 ~]#Ls / usr / SBIN /
I still don't give up and I can't find ifconfig with find command< br />
[ root@centos1 ~]# find / -name " ifconfig"
at this time, I have the bottom of my mind. I should replace ifconfig with a command. Check on the network, sure enough, IP command has been used instead of ifconfig command. The common parameters of IP command are listed below< The code is as follows:
IP [option] operation object {link | addr | route...} & lt/ p> & lt; p># The network interface information is shown in the IP link show, and the network interface information is shown in the network interface information < br /
35; IP link set eth0 UPI 35; opening the network card
35; IP link set eth0 down # IP link set eth0 down # closing the network card < br / # IP link set eth0 promise on # opening the mixed mode of the network card < br /
35; IP link set eth0 Up0 upupi 35; opening the mixed mode of the network card < mixed mode of the network card is opening the mixed mode of the mixed mode of the network card < br / < br /
# IP link set set set set set set the IP link set set eth0 to be set as the network interface to be described in the network card, which is the network interface interface interface and the next one of the network card, which is defined in the network card, which is defined in the network interface interface H0 MTU 1400 ﹥ set the maximum transmission unit of network card
﹥ IP addr show ﹥ display IP information of network card
﹥ IP addr add 192.168.0.1/24 dev eth0 ﹥ set eth0 IP address 192.168.0.1
﹥ IP addr del 192.168.0.1/24 dev eth0 ﹥ delete eth0 IP address & lt/ p> & lt; p># IP route list # view routing information
# IP route add 192.168.4.0/24 via 192.168.0.254 dev eth0 # set the gateway of 192.168.4.0 network segment to 192.168.0.254, Data goes through eth0 interface
# IP route add default via 192.168.0.254 dev eth0 # setting default gateway to 192.168.0.254
# IP route del 192.168.4.0/24 # deleting gateway of 192.168.4.0 network segment
# IP route del default # deleting default route
after entering IP addr command, it is found that enp2s0 network card (this enp2s0 is my network card) has no IP address< br />
[ root@centos1 ~]#IP addr
since there is no IP address, go directly to the / etc / sysconfig / network scripts directory to see the configuration file name of the IP information of the network card< br />
[ root@centos1 ~]# ls /etc/sysconfig/network-scripts/
ifcfg-enp2s0 ifdown-eth ifdown-post ifdown-Team ifup-aliases ifup-ipv6 ifup-post ifup-Team init.ipv6-global
ifcfg-lo ifdown-ippp ifdown-ppp ifdown-TeamPort ifup-bnep ifup-isdn ifup-ppp ifup-TeamPort network-functions
ifdown ifdown-ipv6 ifdown-routes ifdown-tunn el ifup-eth ifup-plip ifup-routes ifup-tunnel network-functions-ipv6
ifdown-bnep ifdown-isdn ifdown-sit ifup ifup-ippp ifup-plusb ifup-sit ifup-wireless
[ root@centos1 ~]#
from the results, the name of the configuration file used to save the IP information of the network card has changed from ifcfg-eth0 to ifcfg-enp2s0. Well, since you gave him such a name, I'll use it first. Cat ifcfg-enp2s0
first[ root@centos1 ~]#Cat / etc / sysconfig / network scripts / ifcfg-enp2s0
code
the code is as follows:
hwaddr = 00: e0:69:01:6a:96
type = Ethernet
bootproto = DHCP
defraute = yes
peerdns = yes
peeroutes = yes
IPv4_ FAILURE_ FATAL=no
IPV6INIT=yes
IPV6_ AUTOCONF=yes
IPV6_ DEFROUTE=yes
IPV6_ PEERDNS=yes
IPV6_ PEERROUTES=yes
IPV6_ FAILURE_ Fat = no
name = enp2s0
UUID = 5b0a7d76-1602-4e19-aee6-29f57618ca01
onboot = no
from the above configuration, we can see that although bootproto = DHCP, onboot = No. here, use VI to change onboot = no to onboot = yes, and then restart CentOS< br />
[ root@centos1 ~]#Shutdown - R
after the restart, enter the account number and password to enter the command prompt operator and continue to view the network card information with IP addr. The results are as follows:
[ root@centos1 ~]# ip add
1: lo: < LOOPBACK,UP,LOWER_ UP> mtu 65536 qdisc noqueue state UNKNOWN
link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
inet 127.0.0.1/8 scope host lo
valid_ lft forever preferred_ lft forever
inet6 ::1/128 scope host
valid_ lft forever preferred_ lft forever
2: enp2s0: < BROADCAST,MULTICAST,UP,LOWER_ UP> mtu 1500 qdisc pfifo_ fast state UP qlen 1000
link/ether 00:e0:69:01:6a:96 brd ff:ff:ff:ff:ff:ff
inet 172.8.1.200/24 brd 172.8.1.255 scope global enp2s0
valid_ lft forever preferred_ lft forever
inet6 fe80::2e0:69ff:fe01:6a96/64 scope link
valid_ lft forever preferred_ lft forever
[ root@centos1 ~]#
from the above results, we can see that the IP address assigned through DHCP is 172.8.1.200. Although it is a testing machine, we should configure a fixed IP for this machine for the convenience of remote connection in the future
open ifcfg-enp2s0 with VI, input the following parameters, and then annotate bootproto = DHCP with #
code
the code is as follows:
ipaddr0 = 172.8.1.211
prefix 0 = 24
gateway 0 = 172.8.1.1
dns1 = 172.8.1.1
the complete parameters are as follows, OK, the network is configured.
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is false
for relevant information,
you can ask the school office and the academic affairs office
if you like,
it's sunny.
is false
for relevant information,
you can ask the school office and the academic affairs office
if you like,
it's sunny.
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