CODEWITHSUNDEEP
shellsed
user766353
user766353

Reputation: 537

sed command doesn't work in shell script

I have a shell script that looks like the following:

#!/bin/bash
for file in $1/*.html; do
   echo "Working on $file ..."

   # delete headers in html files.
   sed -n '1,9d' $file

   # delete the last 4 lines of the file.
   sed -n -e :a -e '1,4!{P;N;D;};N;ba'

   # ant runs regex task on all apt files
   ant -Dtarget.file=$file

   # all .html are changed to .md
   mv $file `echo $file | sed 's/\(.*\.\)html/\1md/'` ;

done

but the script hangs on the first sed command and I have to force exit. I'm wondering if there's anything wrong about the way I've set this up?

If I remove the first two sed commands, I can run the other parts of the script, including that final one with a sed command in it.

Upvotes: 1

Views: 3741

Answers (3)

paxdiablo
paxdiablo

Reputation: 882606

I think you'll find it's hanging on the second sed, where you don't provide a file name to work on. In other words, it will take its standard input from the terminal.

You won't get output from the first sed since -n prevents the default printing of the pattern space. If you're trying to change the actual file rather than output the changes to standard output, you need the -i inplace editor.

If you can't use the inplace option, you'll have to do something like:

sed -n '1,9d' ${file} >${file}.tmp0
sed -n -e :a -e '1,4!{P;N;D;};N;ba' ${file}.tmp0 >${file}
rm -f ${file}.tmp0

This uses a temporary file for making the changes except for the last change, which overwrites the original file. Keep in mind that this changes the original - if that's not what you want, you can change the second sed to output to ${file}.tmp2 but you'll have to adjust your subsequent commands to take that into account.

If you really want to delete the first nine and last four lines of a file, you can also use something like:

head --lines=-4 oldfile.txt | tail --lines=+10 >newfile.txt

provided you have a [rd]ecent implementation of head/tail. The head prints all but the last four lines and the tail starts printing at line ten.

Upvotes: 4

James Polley
James Polley

Reputation: 8181

Change the first line to:

#!/bin/bash -x

This will show you each command as bash executes it, after all expansions have taken place.

My guess is that $1/*.html isn't matching any files; hence the sed line evaluates to just sed -n '1,9d' and so sed is trying to read from stdin

Upvotes: 0

Foo Bah
Foo Bah

Reputation: 26281

What command are you running? If nothing matches the file glob, then it will physically insert the unexpanded string there.

For example, try this:

$ for i in foo/*.bar; do echo $i; done
foo/*.bar

So your sed script is bombing out because the file name it is receiving doesn't exist, which means it will act as if no file argument was specified (ie stdin).

Upvotes: 0

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