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FOR loop problem: Kindly Help

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Top Forums Shell Programming and Scripting FOR loop problem: Kindly Help
# 1  
Old 10-04-2007
FOR loop problem: Kindly Help
Hello Friends,

I have file called reference.txt which looks like below,

Quote:
wahabeua |3.1 |wahabeua-1.1.11 |Not available |Code Coverage
ferrari12 |4.0 |ferrari12-1.1.21|Not applicable |Code Coverage
masterpiece |4.1 |masterpiece-2.2.33 |File Changes |Code Coverage
columns are been formed by delimiter "|" as shown. I am required to scan whole file line by line & compare 1st 2 columns with new name (ex. ferrari) & new version (2.1).

CODE PURPOSE: whnever new project formed compare if it already exists & compare if version updated.

My Code
Quote:
for file in $(cat reference.txt);do
if [[ `echo $file | cut -d ',' -f1` == $PROJECT ]];then
echo "Project Matched"
if [[ `echo $file | cut -d ',' -f2` == $VERSION ]];then
echo "Project & Version Matched"
else
echo "Project Matched but Version Revised"
fi
else
echo "Complete New Project"
fi
echo "$n"
let "n += 1"
done
By executing my code i am finding,
1. its looping for 19 times instead of looping for only 3 times. & providing 3 output for 3 rows.

Kindly help. Thanks a lot.
Last edited by okdev; 10-04-2007 at 08:55 PM..
# 2  
Old 10-04-2007
Perhaps try something like this....
Code:
oldIFS="$IFS"
IFS='|'
while read f1 f2 etc
do
   echo do something with $f1 and $f2
done < reference.txt
IFS="$oldIFS"

# 3  
Old 10-04-2007
Thanks But Keep in touch
Sir thanks a lot for this quick solution. It absolutely gonna help. But stay connected.. if i need some more help!!!

Smilie

Thanks
# 4  
Old 10-05-2007
Continuation Problem
Once i find PROJECT matched & project VERSION matched. I am supposed to update $f3 & $f4 (3rd & 4th field) of same file.

Example:
reference.txt
Quote:
wahabeua |3.1 |wahabeua-1.1.11 |Not available |Code Coverage
ferrari12 |4.0 |ferrari12-1.1.21|Not applicable |Code Coverage
masterpiece |4.1 |masterpiece-2.2.33 |File Changes |Code Coverage
New Project with following variables (I have)
Quote:
PROJECT=ferrari12
VERSION=4.0
LABLE=ferrari12-1.1.25
Reference.txt should look like follwing.

Quote:
wahabeua |3.1 |wahabeua-1.1.11 |Not available |Code Coverage
ferrari12 |4.0 |ferrari12-1.1.25|Not applicable |Code Coverage
masterpiece |4.1 |masterpiece-2.2.33 |File Changes |Code Coverage
Code:
Quote:
#!/bin/sh

PROJECT=ferrari12
VERSION=4.0
LABLE=ferrari12-1.1.25

n=1 # Line Number

oldIFS="$IFS"
IFS='|'
while read f1 f2 f3 etc
do
if [ $f1 == $PROJECT ];then
echo "Line $n: Project Matched"
if [ $f2 == $VERSION ];then
echo " Version Matched"
sed 's/$f3/$LABLE/' # Lable Replaced
else
echo " Version Not Matched"
fi
else
echo "Line $n: New Project"
fi
let "n += 1"
done < reference.txt
IFS="$oldIFS"
some how sed is not working for me there to replace lable. Do you have some logic to fit in this loop.

Please Help! Thanks Smilie
# 5  
Old 10-06-2007
People Kindly Help
People Please Help me out.
ThanksSmilie
# 6  
Old 10-06-2007
Hi.

Use double quotes in this situation -- single quotes do not allow variable expansion:
Code:
sed 's/$f3/$LABLE/' # Lable Replaced

I suggest you look over some tutorials like BASH Programming - Introduction HOW-TO ... cheers, drl
# 7  
Old 10-06-2007
sed "s/$f3/$LABle/"


cheers,
Devaraj Takhellambam
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