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Why can't my QQ always modify my personalized signature?
It may be a network delay or a malicious plug-in in QQ.

The method of testing whether the network is delayed;

Ping local IP.

For example, the local IP address is 172. 168.200.2. Execute the command ping172.168.200.2. If there is nothing wrong with the network card installation configuration, it should be displayed as follows:

replay from 172. 168.200 . 2 bytes = 32 time & lt; 10 ms

ping statistics for 172. 168.200 . 2

Packet sent = 4 received = 4 lost = 00% lost.

About a millionth of a second.

Minimum value = 0maxiumu =1msaverage = 0ms.

If the command is executed in MS-DOS mode, the display content is: Requesttimedout, indicating that there is a problem with the installation or configuration of the network card. Disconnect the network cable and execute the command again. If the display is normal, it means that the IP address used by this machine may be duplicate with the IP address of another machine in use. If it is still abnormal, it means that there is something wrong with the installation or configuration of the local network card, and it is necessary to continue to check the relevant network configuration.

Ping gateway IP.

Assume that the gateway IP is:192.168.1.12, and execute the command ping192.168.1. When executing this command in MS-DOS mode, if the following information is displayed:

reply from 192. 168. 1. 12 bytes = 32 time & lt; 1msTTL=64

ping statistics for 192. 168. 1. 12

PacketsSent=4Received=4Lost=0

About a millionth of a second.

Minimum value = 1 ms Maximum value = 9 ms Average time = 5 ms.

Indicates that the gateway router in the LAN is operating normally. On the contrary, there is a problem with the gateway.

Ping remote IP.

This command can detect whether the computer can access the Internet normally. For example, the IP address of a local telecom operator is: 202.438+002.48.6438+0. Execute the command in MS-DOS mode: ping 202.102.48.141. If the screen shows:

reply from 202. 102 . 48 . 14 1 bytes = 32 time = 33 msttl = 252

reply from 202. 102 . 48 . 14 1 bytes = 32 time = 2 1 msttl = 252

reply from 202. 102 . 48 . 14 1 bytes = 32 time = 5 msttl = 252

reply from 202. 102 . 48 . 14 1 bytes = 32 time = 6 msttl = 252

ping statistics for 202. 102.48 . 14 1

Packet sent = 4 received = 4 lost = 00% lost.

About a millionth of a second.

Minimum value = 5ms Maximum value = 33ms Average time =16ms

It means that the operation is normal and you can surf the Internet normally. Otherwise, there is a problem with the host file (windows/host).