怎么做：`如果ls / etc / * release 1> / dev / null 2>＆1`工作？ 请解释一下

Question

Could someone help me under stand the condition ls /etc/*release 1>/dev/null 2>&1 that's contained in the code:

    if ls /etc/*release 1>/dev/null 2>&1; then
       echo "<h2>System release info</h2>"
       echo "<pre>"
       for i in /etc/*release; do

           # Since we can't be sure of the
           # length of the file, only
           # display the first line.

           head -n 1 $i
       done
       uname -orp
       echo "</pre>"
    fi

I pretty much don't understand any of that line but specifically what I wanted to know was:

1. Why dose it not have to use the 'test' syntax ie [ expression ] ?
2. The spacing in the condition also confuses, is 1>/dev/null a variable in the ls statement?
3. what is 2>&1 ?

I understand the purpose of this statement, which is; if there exists a file with release in it's name under the /etc/ directory the statement will continue, I just don't understand how this achieves this.

Thanks for you help

Answer 1

[ isn't a special character, it's a command ( /bin/[ or /usr/bin/[ , usually a link to test ). That means

if [ ...
if test ...

are the same. For this to work, test ignores ] as last argument if it's being called [ .

if simply responds to the exit code of the command it invokes. An exit code of 0 means success or "true".

1>/dev/null 2>&1 redirects stdout ( 1 ) to the device /dev/null and then stderr ( 2 ) to stdout which means the command can't display and output or errors on the terminal.

Since stdout isn't a normal file or device, you have to use >& for the redirection.

At first glance, one would think that if [ -e /etc/*release ] would be a better solution but test -e doesn't work with patterns.

Answer 2

The test programm just evaluate its arguments and return a code 0 or 1 to tell whether it was true or not.

But you can use any shell commands/function with a if . It will do the then part if the return code ( $? ) was 0. So, here, we look if ls return 0 (a file matched), or not.

So, in the end, it's equivalent to write if [ -e /etc/*release ] ; then if [ -e /etc/*release ] ; then , which is more "shell-liked".

The last two statements 1>/dev/null and 2>&1 are just here to avoid displaying the output of the ls

• 1>/dev/null redirect stdout to /dev/null , so the standard out is not shown
• 2>&1 redirect stderr to stdout. Here, stdout is redirected to /dev/null , so everything is redirected to /dev/null