rm fails to delete files by wildcard from a script

2019-03-08 10:00发布

I've run into a really silly problem with a Linux shell script. I want to delete all files with the extension ".bz2" in a directory. In the script I call

rm "$archivedir/*.bz2"

where $archivedir is a directory path. Should be pretty simple, shouldn't it? Somehow, it manages to fail with this error:

rm: cannot remove `/var/archives/monthly/April/*.bz2': No such file or directory

But there is a file in that directory called test.bz2 and if I change my script to

echo rm "$archivedir/*.bz2"

and copy/paste the output of that line into a terminal window the file is removed successfully. What am I doing wrong?

标签: linux bash shell
6条回答
ゆ 、 Hurt°
2楼-- · 2019-03-08 10:01

Your original line

rm "$archivedir/*.bz2"

Can be re-written as

rm "$archivedir"/*.bz2

to achieve the same effect. The wildcard expansion is not taking place properly in your existing setup. By shifting the double-quote to the "front" of the file path (which is legitimate) you avoid this.

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Lonely孤独者°
3楼-- · 2019-03-08 10:06

The quotes are causing the string to be interpreted as a string literal, try removing them.

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一夜七次
4楼-- · 2019-03-08 10:15

Why not just rm -rf */*.bz2? Works for me on OSX.

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Bombasti
5楼-- · 2019-03-08 10:17

Just to expand on this a bit, bash has fairly complicated rules for dealing with metacharacters in quotes. In general

  • almost nothing is interpreted in single-quotes:

     echo '$foo/*.c'                  => $foo/*.c
     echo '\\*'                       => \\*
    
  • shell substitution is done inside double quotes, but file metacharacters aren't expanded:

     FOO=hello; echo "$foo/*.c"       => hello/*.c
    
  • everything inside backquotes is passed to the subshell which interprets them. A shell variable that is not exported doesn't get defined in the subshell. So, the first command echoes blank, but the second and third echo "bye":

    BAR=bye echo `echo $BAR`
    BAR=bye; echo `echo $BAR`
    export BAR=bye; echo `echo $BAR`
    

(And getting this to print the way you want it in SO takes several tries is apparently impossible...)

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Juvenile、少年°
6楼-- · 2019-03-08 10:20

TL;DR

Quote only the variable, not the whole expected path with the wildcard

rm "$archivedir"/*.bz2

Explanation

  • In Unix, programs generally do not interpret wildcards themselves. The shell interprets unquoted wildcards, and replaces each wildcard argument with a list of matching file names. if $archivedir might contain spaces, then rm $archivedir/*.bz2 might not do what you

  • You can disable this process by quoting the wildcard character, using double or single quotes, or a backslash before it. However, that's not what you want here - you do want the wildcard expanded to the list of files that it matches.

  • Be careful about writing rm $archivedir/*.bz2 (without quotes). The word splitting (i.e., breaking the command line up into arguments) happens after $archivedir is substituted. So if $archivedir contains spaces, then you'll get extra arguments that you weren't intending. Say archivedir is /var/archives/monthly/April to June. Then you'll get the equivalent of writing rm /var/archives/monthly/April to June/*.bz2, which tries to delete the files "/var/archives/monthly/April", "to", and all files matching "June/*.bz2", which isn't what you want.

The correct solution is to write:

rm "$archivedir"/*.bz2
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做自己的国王
7楼-- · 2019-03-08 10:26

I've seen similar errors when calling a shell script like ./shell_script.sh from another shell script. This can be fixed by invoking it as sh shell_script.sh

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