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. 0 7 Answers Something...
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
James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.
