Do not start loop if there is no files in director

Question

All,

I am running BASH in Solaris 10

I have the following shell script that loops in a directory depending on the presence of CSV files.

The problem is with this piece of code is that it still does one loop even if there is no CSV files in that directory and then calls SQL loader.

SQLLoader then produces a log file because there is no file to process and this is beginning to mess up my directory filling it with log files.

for file in *.csv ; 
do 
    echo "SQLLoader is reading : " $file
    sqlldr <User>/<Password>@<DBURL>:<PORT>/<SID> control=sqlloader.ctl log=$inbox/$file.log data=$inbox/$file
done

How do I stop it going into a loop if there is no CSV files in that directory of $inbox

Answer 1

Use find to search files

for file in `find -name "*.csv"` ;

Answer 2

First off, using nullglob is the correct answer if it is available. However, a POSIX-compliant option is available.

The pattern will be treated as literal text if there are no matches. You can catch this with a small hack:

for file in *.csv; do
    [ -f "$file" ] || break
    ...
done

When there are no matches, file will be set to the literal string *.csv, which is not the name of a file, so -f "$file" will fail. Otherwise, file will be set in turn to the name of each file matching the pattern, and -f "$file" will succeed every time. Note this will work even if there is an file named *.csv. The drawback is that you have to make a redundant test for each existing file.

Answer 3

Say:

shopt -s nullglob

before your for loop.

This is not the default, and saying for file in *.csv when you don't have any matching files expands it to *.csv.

Quoting from the documentation:

nullglob

If set, Bash allows filename patterns which match no files to expand to a null string, rather than themselves.