UNIX for Dummies Questions & Answers

Hi All,

I'm trying to run a simple shell program to change all the files named *.cvs to *.txt. I am trying to use WHILE and this is what I have so far:
This changes the first file from *.cvs to *.txt, but it is not cycling through the other files.

My suspicion is that I don't have the right WHILE [CONDITION], but since I haven't used WHILE before I'm not sure how to do something like while [ $filename!=*.txt ] (psuedo code). Also, I am by no means tied to the WHILE command, if someone knows another command to do this I'd like to know.

Cheers,
ScKaSx

Code:

#!/bin/ksh
for i in *.cvs
do
  mv $file ${file%.lis}".txt"
done

Hi Jim and Everyone Else,

I should've mentioned but I'm using the bash shell so I get errors when I try your solution. What would be the bash equivalent? Also, is there a way to do it with WHILE (just out of curiousity)?

Cheers,
ScKaSx

try this in bash
for file in *.cvs ; do echo mv $file `basename $file .cvs`.txt ; done

if the commands print out correctly, remove the echo

I never use while loops in sh (or bash), but I see nothing that changes your filename, so I don't see where your loop would exit.

Yes, there is. The basic operation is to read a stream of data one line at a time and assign the content of the line to a variable:

Code:

while read var ; do
     echo $var
     do_something_else -with $var
done

You can also split the line into several pieces if you name several variables afer the read statement. The splitting is done at word boundaries (white space). If you have fewer columns than variables the last variables are empty, if you have fewer variables than columns the last variable gets the remainder of the line:

Code:

while read var1 var2 var3 ; do
     echo "first part is  : $var1"
     echo "second part is : $var2"
     echo "rest of line is: $var3"
done

So the only thing left is to create the data stream to pour it into the loop. This is done with a "pipeline":

Code:

ls -1 *csv | while read filename ; do
     mv ${filename}.txt
done

This would rename "a.csv" to "a.csv.txt" which is not exactly what we want. We have to first cut off the ".csv" and only then add ".txt":

Code:

ls -1 *csv | while read filename ; do
     mv ${filename%%.*}.txt
done

To show you an example for what a loop with multiple variables can do: You want a list of filenames like in "ls -l" but only the filenames and the sizes. Lets examine the output of "ls -l" :

Code:

# ls -l
total 160
-rw-r--r--   1 ehow     aixteam        3077 Nov 29 12:24 gedas.err
-rw-r--r--   1 ehow     aixteam       25038 Dec  7 10:47 gedas.log
-rw-r--r--   1 ehow     aixteam       11658 Feb 24 16:51 smit.log
-rw-r--r--   1 ehow     aixteam        3829 Feb 24 16:51 smit.script

We only want the 5th and the 9th column. most people would (mis-)use awk for that, but its a lot simpler:

Code:

ls -l | while read junk junk junk junk fsize junk junk junk fname ; do
     echo "${fname}\t${fsize}"
done

I hope this helps.

bakunin

Hi Guys,

Thanks alot! Scott1256ca, the for command worked brillantly! Also thanks for the tutorial on while bakunin!

Cheers,
ScKaSx