How can I use the test command for an arbitrary number of files, passed in using an argument with a wildcard?
For example:
test -f /var/log/apache2/access.log.* && echo "exists one or more files"
Currently, it prints
error: bash: test: too many arguments
This solution seems to me more intuitive:
if [ `ls -1 /var/log/apache2/access.log.* 2>/dev/null | wc -l ` -gt 0 ];
then
echo "ok"
else
echo "ko"
fi
stderr to a null file using 2>/dev/null since if no match found, the error of ls wouldn't be printed out.To avoid "too many arguments error", you need xargs. Unfortunately, test -f doesn't support multiple files. The following one-liner should work:
for i in /var/log/apache2/access.log.*; do test -f "$i" && echo "exists one or more files" && break; done
By the way, /var/log/apache2/access.log.* is called shell-globbing, not regexp. Please see Confusion with shell-globbing wildcards and Regex for more information.
First, store files in the directory as an array:
logfiles=(/var/log/apache2/access.log.*)
Then perform a test on the count of the array:
if [[ ${#logfiles[@]} -gt 0 ]]; then
echo 'At least one file found'
fi
['/var/log/apache2/access.log.*'] (one element containing the '*'). Suggestion: [[ ${#logfiles[@]} -gt 1 || -e ${logfiles[0] ]] or simply [[ -e ${logfiles[0] ]] (as dmaticzka does)
shopt -s nullglob to avoid having an element in the array content if no file exists. In that case, the variable will be empty and the following test easier to make.
if [[ ${#logfiles[@]} -gt 0 ]] && [[ -f ${#logfiles[@]} ]] ; then . Sorry for being late for the party, so the shopt -s nullglob will not be necessary?
if [[ ${#logfiles[@]} -gt 0 ]] && [[ -e ${logfiles[@]} ]]; then to catch the condition for 0 files! Or use -f or -d to test for just files/directories ;).This one is suitable for use with the Unofficial Bash Strict Mode, no has non-zero exit status when no files are found.
The array logfiles=(/var/log/apache2/access.log.*) will always contain at least the unexpanded glob, so one can simply test for existence of the first element:
logfiles=(/var/log/apache2/access.log.*)
if [[ -f ${logfiles[0]} ]]
then
echo 'At least one file found'
else
echo 'No file found'
fi
If you wanted a list of files to process as a batch, as opposed to doing a separate action for each file, you could use find, store the results in a variable, and then check if the variable was not empty. For example, I use the following to compile all the .java files in a source directory.
SRC=`find src -name "*.java"`
if [ ! -z $SRC ]; then
javac -classpath $CLASSPATH -d obj $SRC
# stop if compilation fails
if [ $? != 0 ]; then exit; fi
fi
You just need to test if ls has something to list:
ls /var/log/apache2/access.log.* >/dev/null 2>&1 && echo "exists one or more files"
Variation on a theme:
if ls /var/log/apache2/access.log.* >/dev/null 2>&1
then
echo 'At least one file found'
else
echo 'No file found'
fi
ls -1 /var/log/apache2/access.log.* | grep . && echo "One or more files exist."
Or using find
if [ $(find /var/log/apache2/ -type f -name "access.log.*" | wc -l) -gt 0 ]; then
echo "ok"
else
echo "ko"
fi
-maxdepth 1 before -type f to not go into subdirectories. For speedup, use this: find /var/log/apache2/ -maxdepth 1 -type f -name "access.log.*" -print -quit (quit after first match)This condition below doesn't produce stderr. the condition's blackhole (/dev/null) doesn't prevent the stderr in cmd.
if [[ $(ls -1 /var/log/apache2/access.log.* | wc -l ) -gt 0 ]] 2> /dev/null
therefore I suggests this code.
if [[ $(ls -1 /var/log/apache2/access.log.* | wc -l ) -gt 0 ]] 2> /dev/null
then
echo "exists one or more files."
fi
more simplyfied:
if ls /var/log/apache2/access.log.* 2>/dev/null 1>&2; then
echo "ok"
else
echo "ko"
fi
Faced the same issue and I put together one-liner using test, without error if no such file exists:
[ $(ls /var/log/apache2/access.log.* 2>/dev/null | wc -l) -gt 0 ] && echo "exists one or more files" || echo "no log files"
