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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.

2010/11/12
1
158
11/12/2010 8:43:29 AM

Accepted Answer

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
2010/11/12
235
11/12/2010 8:35:33 AM

"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/

2015/01/31

Here is a while loop:

while read filename
do
    echo "Printing: $filename"
    cat "$filename"
done < filenames.txt
2010/11/12

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> _
2010/11/12

cat `cat filenames.txt`

will do the trick

2017/03/21

Source: https://stackoverflow.com/questions/4162821
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