remove part of a line

Shell Programming and Scripting
1

Hi
I need some help with using shell script to analyze the content of a file. I hope someone can help me.

I have a file with content like the following:
/foldera/database/procedure/a.proc$$/version1/2
/folderb/database/procedure/proj1/b.proc$$/version2/2

I need to write a shell script to get the lines that contains ".proc" but I need to delete anything after .proc.
For example, for the above two lines, I need to get
/foldera/database/procedure/a.proc
/folderb/database/procedure/proj1/b.proc

So that I can go to each file and perform more analysis

when I do

grep ".proc" file.txt

I get the entire line including $$...
I am not sure how to remove the $$.. part

Please help

Thanks

2

someone post the similar problem as mine.

It is to remove .null T from 'M000071.null T'
The script

echo 'M00071.null T' | sed 's#^\(M[0-9][0-9]*\)[.]null.*#\1#'

works

I try

echo '/foldera/database/procedure/a.proc@@/version1/2' | sed 's#^\(*\)[.]proc.*#\1#'

My script just return the whole line.

I can't figure out what's wrong with my script...
Thanks in advance for any help

3

Why don't you simply do this:

$ cat file.txt
/foldera/database/procedure/a.proc$$/version1/2
/folderb/database/procedure/proj1/b.proc$$/version2/2
$ cut -d'$' -f1 < file.txt
/foldera/database/procedure/a.proc
/folderb/database/procedure/proj1/b.proc

HTH

4 
 cat file.txt | awk -F '\\$\\$' '{print $1}'
5

And another one:

or:

6

Hi All
Thank you all for your help. I am using cut, because it's easier to understand for me

I have an issue with grep before pipe the result to get the $$.. removed.

I have
grep ".proc" file.txt | cut ....

From my understanding, it should only return lines with .proc
However, I am not getting how come the following line also gets returned?

/vobs/aw/database/proc/Account$$/main

Can someone help me, how can I modify my grep so that only something like a.proc or b.proc get returned ?

Thanks

7 
sed -n -e "s/\$\$.*//p" input.txt
8

Hi Vino
I tried your code. However, it doesn't solve my grep problem.

Do you have any other solution?

Thanks

9

cat file | grep proc | awk -F"$" '{ print $1 }'

10

Hi All
I have just realized that beside removing anything after $$, I also need to remove anything before the file name.

That is, if my file has the following:
/foldera/database/procedure/a.proc$$/version1/2
/folderb/database/procedure/proj1/b.proc$$/version2/2

I hope to have the following result:
a.proc
b.proc

I have tried using unix "cut" command, but I am no where getting the result I wanted... could someone please give me some example...I am really stucked.

Thanks

11

That's because the "." stands for any character and grep matches it to the slash that comes before proc (/proc). To match "." escape it so that it loses its special meaning...

grep "\.proc" file.txt | cut
12

What is your grep problem ?

If your input is

/vobs/aw/database/proc/Account$$/main
/foldera/database/procedure/a.proc$$/version1/2
/folderb/database/procedure/proj1/b.proc$$/version2/2

then the expected output is

a.proc
b.proc

Right ?

Try this

sed -n -e "s/.*\/\([^\/]*\.proc\)\$\$.*/\1/p" input.txt
13

cat file | cut -d "$" -f1

14

Can u try this one out........

while read a
do
echo $a | awk -F"." '{print $1".proc"}'
done < filename

15

This looks simpler

sed -ne "s/\$\$.*//p" input.txt | xargs basename
16

An awk:

awk '/\.proc/{sub(/\$.*/,"",$0);print}' RS='/' file.txt
17

let
cat dot.txt
/foldera/database/procedure/a.proc$$/version1/2
/folderb/database/procedure/proj1/b.proc$$/version2/2
/folderb/database/procedure$$/proj1/version3/45
$
$ cat dot.txt | awk -F"." '/.proc\$\$/ {print $1".proc"}'
/foldera/database/procedure/a.proc
/folderb/database/procedure/proj1/b.proc