Bash - Getting a for loop to process variables with a space in them

davee · 10-04-2005, 09:09 AM

Say I have a file:
[user@myserver conf]$ cat schedule.txt
051004,echo "helloworld",100
051004,echo "helloworld",100
051004,echo "helloworld",100
051004,echo "helloworld",100

How can I run each command in the list from one line?

I've tried:
[user@myserver conf]$ for i in `awk -F, '{print $2}' schedule.txt`; do `$i`; done
bash: "helloworld": command not found
bash: "helloworld": command not found
bash: "helloworld": command not found
bash: "helloworld": command not found

Which runs the echos, but thinks the argument is another command...

I suppose my real question is - How can I get the for loop to treat each line as a single command/variable?

Dave

druuna · 10-04-2005, 09:36 AM

Hi,

Im'n not entirely sure what it is you actually want, but maybe this wil help:

Code:

#!/bin/bash

# put whole line in variable
cat schedule.txt | while read THISLINE
do
  echo $THISLINE
done

# only the echo "helloworld" part
awk -F"," '{print $2}' schedule.txt

I do believe you are asking for the first one (the while read construct).

Anyway, hope this helps.

anomie · 10-04-2005, 09:38 AM

Here's an example that should answer your question.

Code:

awk '{ print; }' sample | sh

homey · 10-04-2005, 10:46 AM

Do you need a loop?
awk -F, '{print $2}' schedule.txt

If you do want to use a loop, maybe this will get you started...

Code:

#!/bin/bash
# Example: ./test schedule.txt
cat $1 | \
while read i; do
awk -F, '{print $2}'
done

davee · 10-08-2005, 11:21 AM

Thanks folks - all these are helpful; solved my problem.

Dave