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SED - replace only on part of the string

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Top Forums Shell Programming and Scripting SED - replace only on part of the string
# 8  
Old 01-14-2012
See if awk works for you:
Code:
awk -F\) '{gsub(/,/,"|",$1)}1' RS=\( ORS=\( OFS=\)

Quote:
Originally Posted by Shell_Life
Extracted from the 'sed' book:
\b is not standard sed, but GNU sed only
This User Gave Thanks to Scrutinizer For This Post:
Sephiburp
# 9  
Old 01-14-2012
Quote:
Originally Posted by Scrutinizer
See if awk works for you:
Code:
awk -F\) '{gsub(/,/,"|",$1)}1' RS=\( ORS=\( OFS=\)

It works !

But I don't understand all the parts.

First I do some change so that it's easier to understand :
Code:
awk '{gsub(",","|",$1)}1' FS=\) RS=\( OFS=\) ORS=\(

Well I understand that we separate each record with '(' and each field with ')'.
So in the first field of each record we are sure that all comma must be replaced by pipe.

But I don't understand the '1' in {gsub(",","|",$1)}1


_________________________________________________________________________________
EDIT

In fact there is one thing to fix : if the last character of a line is not a ')', the next line is merged :
Code:
$ cat file
PARMA1=('VAL11'),PARMA2=(VAL21 ), PARMA3= VAL31 ,PARMA4= VAL41 ,PARMA5 = 'VAL51',PARMA6=(VAL61)
PARMB1=('VAL11'),PARMB2=(VAL21 , VAL22,VAL23), PARMB3= VAL31 ,PARMB4=(VAL41,VAL42),PARMB5 = 'VAL51',PARMB6=(VAL61)
PARMC1=('VAL11'),PARMC2=(VAL21 ), PARMC3= VAL31 ,PARMC4= VAL41 ,PARMC5 = 'VAL51',PARMC6=(VAL61)
PARMD1=('VAL11'),PARMD2=(VAL21 , VAL22,VAL23), PARMD3= VAL31 ,PARMD4=(VAL41,VAL42),PARMD5 = 'VAL51',PARMD6='VAL61'
PARME1=('VAL11'),PARME2=(VAL21 ), PARME3= VAL31 ,PARME4= VAL41 ,PARME5 = 'VAL51',PARME6=(VAL61)

$ cat file | awk '{gsub(",","|",$1)}1' FS=\) RS=\( OFS=\) ORS=\(
PARMA1=('VAL11'),PARMA2=(VAL21 ), PARMA3= VAL31 ,PARMA4= VAL41 ,PARMA5 = 'VAL51',PARMA6=(VAL61)
PARMB1=('VAL11'),PARMB2=(VAL21 | VAL22|VAL23), PARMB3= VAL31 ,PARMB4=(VAL41|VAL42),PARMB5 = 'VAL51',PARMB6=(VAL61)
PARMC1=('VAL11'),PARMC2=(VAL21 ), PARMC3= VAL31 ,PARMC4= VAL41 ,PARMC5 = 'VAL51',PARMC6=(VAL61)
PARMD1=('VAL11'),PARMD2=(VAL21 | VAL22|VAL23), PARMD3= VAL31 ,PARMD4=(VAL41|VAL42),PARMD5 = 'VAL51',PARMD6='VAL61')PARME1=('VAL11'),PARME2=(VAL21 ),.........

Sephi.
Last edited by Sephiburp; 01-14-2012 at 06:15 AM..
# 10  
Old 01-14-2012
Hi, 1 effectuates "print record". A value of 1 outside the brackets makes awk invoke the default action, which is {print $0}.

Strange I cannot reproduce this effect. I get:

Code:
PARMA1=('VAL11'),PARMA2=(VAL21 ), PARMA3= VAL31 ,PARMA4= VAL41 ,PARMA5 = 'VAL51',PARMA6=(VAL61)
PARMB1=('VAL11'),PARMB2=(VAL21 | VAL22|VAL23), PARMB3= VAL31 ,PARMB4=(VAL41|VAL42),PARMB5 = 'VAL51',PARMB6=(VAL61)
PARMC1=('VAL11'),PARMC2=(VAL21 ), PARMC3= VAL31 ,PARMC4= VAL41 ,PARMC5 = 'VAL51',PARMC6=(VAL61)
PARMD1=('VAL11'),PARMD2=(VAL21 | VAL22|VAL23), PARMD3= VAL31 ,PARMD4=(VAL41|VAL42),PARMD5 = 'VAL51',PARMD6='VAL61'
PARME1=('VAL11'),PARME2=(VAL21 ), PARME3= VAL31 ,PARME4= VAL41 ,PARME5 = 'VAL51',PARME6=(VAL61)

Are you on Solaris? If so use nawk or /usr/xpg4/bin/awk
This User Gave Thanks to Scrutinizer For This Post:
Sephiburp
# 11  
Old 01-14-2012
I'm on Linux Suse.
I tried with nawk but it's the same result.
Then I tried with gawk and it works ! Smilie
Thank you.

Sephi.
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