Bash foreach loop

I have an input (let's say a file). On each line there is a file name. How can I read this file and display the content for each one.

7 Answers

Something like this would do:

xargs cat <filenames.txt 

The xargs program reads its standard input, and for each line of input runs the cat program with the input lines as argument(s).

If you really want to do this in a loop, you can:

for fn in `cat filenames.txt`; do echo "the next file is $fn" cat $fn done 

"foreach" is not the name for bash. It is simply "for". You can do things in one line only like:

for fn in `cat filenames.txt`; do cat "$fn"; done 

Reference:

Here is a while loop:

while read filename do echo "Printing: $filename" cat "$filename" done < filenames.txt 

xargs --arg-file inputfile cat 

This will output the filename followed by the file's contents:

xargs --arg-file inputfile -I % sh -c "echo %; cat %" 

You'll probably want to handle spaces in your file names, abhorrent though they are :-)

So I would opt initially for something like:

pax> cat qq.in normalfile.txt file with spaces.doc pax> sed 's/ /\\ /g' qq.in | xargs -n 1 cat <<contents of 'normalfile.txt'>> <<contents of 'file with spaces.doc'>> pax> _ 

If they all have the same extension (for example .jpg), you can use this:

for picture in *.jpg ; do echo "the next file is $picture" done 

(This solution also works if the filename has spaces)

cat `cat filenames.txt` 

will do the trick