Deleting part of a string : string manipulation


 
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# 1  
Old 04-20-2012
Deleting part of a string : string manipulation

i have something like this...

Code:
 echo "teCertificateId" | awk -F'Id' '{ print $1 }' |  awk -F'te' '{ print $2 }'
Certifica

the awk should remove 'te' only if it is present at the start of the string.. anywhere else it should ignore it.

expected output is
Code:
Certificate

# 2  
Old 04-20-2012
Without external programs (fast)
Code:
var=teCertificateId
var2=${var#te}
var3=${var2%Id}
echo "$var3"

With external programs:
Code:
echo "$var" | sed 's/^te//;s/Id$//'

These 2 Users Gave Thanks to Scrutinizer For This Post:
# 3  
Old 04-20-2012
Hi


Code:
$  echo "teCertificateId" | awk -F'Id' '{sub(/^te/,"",$1); print $1 }'
Certificate

Guru,
This User Gave Thanks to guruprasadpr For This Post:
# 4  
Old 04-20-2012
thanks a lot :-)
# 5  
Old 04-20-2012
And a third way:
Code:
echo "teCertificateId"| sed 's/^te//; s/Id$//'

This User Gave Thanks to zaxxon For This Post:
# 6  
Old 04-20-2012
Code:
 
echo "teCertificateId" | nawk '{gsub("^te|Id$","",$0);print}' 
Certificate

# 7  
Old 04-20-2012
A neater way within ksh is to slice the variable and it doesn't spawn a new process. If you know that "te" is to be removed from the begginning and "Id" from the end then it's simply like Scrutinizer wrote:-
Code:
var="teCertificateId"
var2="${var#te}"
echo "${var2%Id}"

however if you just want to trim any two leawding and trailing characters you need to use the wildcard ?:-
Code:
var=teCertificateId
var2="{var##??}"
echo "{var2%%??}"


Does this help?




Robin
Liverpool/Blackburn
UK
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