Is there a way to supress the command output for git if I use backticks in my shell script? Here is my current code:

OUT=$(git status > /dev/null)

Thanks :)

有帮助吗?

解决方案

I think what you're wanting is to suppress the stderr but not the stdout since you still want the value. You could do this instead:

OUT=$(git status 2>/dev/null)

其他提示

In case some output is going to standard error:

OUT=$(git status > /dev/null 2>&1; echo $?)

Of course, this does leave open the question: what is it you want to capture in OUT?

[EDIT] The above will put the return code of git into $OUT.

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