JIYIK CN >

? is a special variable. This variable represents the return value of the previous command. That is to say, when we run certain commands, these commands will return a code after running. Generally, if the command is successfully run, the return value is 0. If an error occurs during the running process, an error code will be returned (this code indicates the reason for the error under the command). It is usually a non-zero number.

We know that to call a variable in the shell, you need to add $ in front of the variable. This is what we will use in the following examples.

Example 1

# test –e /tmp/a.txt; echo $?
0 // /tmp/a.txt exists, so the returned value is 0

For example, if /tmp/a.txt does not exist, the result is non-0. If we actually run the above command, we will find that if /tmp/a.txt does not exist, the return value is 1.

Let's look at another example about $?

Example 2

# echo 'onmpw'
onmpw //Command runs normally
# echo $?
0 //Output is 0
# 1name=onmpw
-bash: 1name=onmpw: command not found //Command runs incorrectly
# echo $?
127 //Output is 127 A non-zero number, indicating that an error occurred in the previous command run
# echo $?
0 //The run result is 0 again This happens because? It indicates the return value of the [previous command].
             // Let's see if its previous command is echo $? This command is executed correctly and outputs //127 So we will return 0 in the next result.

OK, after introducing $?, let's see where it is mainly used. We know that in Linux, there is a dependency between two commands. And the most important thing about this dependency is whether the result of the previous command is correct. At this time, we need to use the command return value we mentioned above. Of course, in addition to the command return value, we also need to use && and || to help.

See the following example

Example 3

# test –e /tmp/a.txt && echo 'Yes' || echo 'No'
Yes ​​//Because /tmp/a.txt exists, the result is Yes

Because /tmp/a.txt exists, the first command runs normally, so && will continue to execute the next command and output Yes, and the echo 'Yes' command is also executed correctly, so the previous results of || are correct and will not be executed further.

&& returns an error value as long as one of the expressions on both sides is wrong. If the previous expression is wrong, the whole expression is considered wrong and the following commands will not be executed. If the previous expression is executed correctly, the following commands will be executed.

|| As long as one of the expressions on both sides is correct, the whole expression will return a correct value. If the previous command is executed correctly, the following command will not be executed.

In example 3, /tmp/a.txt exists, so we replace it with a non-existent file /tmp/b.txt

Example 4

# test –e /tmp/b.txt && echo “Yes” || echo “No”
No //Result output No

Because /tmp/b.txt does not exist, the first command returns an error value. Therefore, && as a whole is wrong. However, the return value before || is wrong, so the subsequent command will continue to be executed, so No is output.

The above text is a bit long-winded. Actually, if we have learned other programming languages, it is easy to understand && and ||. This article mainly introduces the special variable $?, because it is often encountered in shell script programming. I hope this article is helpful to you.