Bash foreach loop

Question

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.

Answer 1

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

Answer 2

"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: http://www.cyberciti.biz/faq/linux-unix-bash-for-loop-one-line-command/

Answer 3

Here is a while loop:

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

Answer 4

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 %"

Answer 5

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> _

Answer 6

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)

Answer 7

cat `cat filenames.txt`

will do the trick