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find single quote in a string and replace it

 
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# 1  
Old 11-03-2010
find single quote in a string and replace it
Hi,

I have variable inside shell script - from_item.
from_item = 40.1'1/16

i have to first find out whether FROM_ITEM contains single quote(').
If yes, then that need to be replace with two quotes ('').

How to do it inside shell script? Please note that inside shell script........
# 2  
Old 11-03-2010
Code:
#!/bin/ksh

FROM_ITEM="40.1'1/16"

echo "OLD:  ${FROM_ITEM}" 

#
# If you actually want two single quotes.
#
SINGLE_QUOTE=$(echo ${FROM_ITEM} | sed "s/'/''/")

#
# If you actually want one double quote.
#
DOUBLE_QUOTE=$(echo ${FROM_ITEM} | sed "s/'/\"/")

echo "SINGLE QUOTE:  ${SINGLE_QUOTE}"

echo "DOUBLE QUOTE:  ${DOUBLE_QUOTE}"


jsmith:..ksh $ ./tsed.ksh  
OLD:  40.1'1/16
SINGLE QUOTE:  40.1''1/16
DOUBLE QUOTE:  40.1"1/16

Last edited by jsmithstl; 11-03-2010 at 07:50 AM.. Reason: Changed code to include double quote example.
# 3  
Old 11-03-2010
thanks jsmithstl for reply.

I have used this already and works fine at unix prompt.

When used same inside shell script , gives following error. can u please help?

Error on line 218:
(SQR 3706) Missing Comma.
from_item=$(echo $from_item | sed "s/'/''/g")

Error on line 218:
(SQR 3710) Unknown argument type.
from_item=$(echo $from_item | sed "s/'/''/g")
# 4  
Old 11-03-2010
Can you post your entire script and what shell you are using?
# 5  
Old 11-04-2010
You can do something like this:
Code:
case $from_item in
  *\'*) echo "$from_item contains a single quote"
        echo "here it is gone: ${from_item%\'*}${from_item#*\'}" 
        echo "here ${from_item%\'*}\"${from_item#*\'} has a double quote"
esac

 
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