BASH Program to remove files, Integer expression expected

Question

#
# if MAXFILES is not set, set to 10
#

if [ -z "MAXFILES" ]
then
        MAXFILES=10
fi

#
# now check to see if the number of files being removed is > MAXFILES
# but only if MAXFILES = 0
#

if [ $# -gt "$MAXFILES" -a "$MAXFILES" -ne 0 ]
then
        # if it is, prompt user before removing files
        echo "Remove $# files (y/n)? \c"
        read reply
        if [ "$reply" = y ]
        then
                rm "$@"
        else
                echo "files not removed"
        fi
else
        # number of args <= MAXFILES
        rm "$@"
fi

The above program I have to remove files. However when I attempt to run it, its telling me that

line 15: [: : integer expression expected

You can assign variable in easy way: if [ $# -gt "${MAXFILES:=10}" -a "$MAXFILES" -ne 0 ] is enough without above lines — Costas
– Costas, Commented Dec 2, 2014 at 16:45

DopeGhoti · Accepted Answer · 2014-12-02 17:39:29Z

The problem is here:

if [ -z "MAXFILES" ]
then
        MAXFILES=10
fi
# ...
if [ $# -gt "$MAXFILES" -a "$MAXFILES" -ne 0 ]

You are checking to see whether the string MAXFILES is zero. Since it's not, $MAXFILES never gets set, and so your later test is lexing as:

if [ $# -gt "" -a "" -ne 0 ]

This is why it's complaining about needing an integer.

What you want to do is this:

if [[ -z "$MAXFILES" ]]; then
  MAXFILES=10
fi

The difference is the dollar sign, which you are missing. Also, I tend to use [[ ]] rather than [ ] for tests (when using bash) as they have some nice built-in sanity checks and are more versatile than [ and test.

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