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find and replace with variable -sed

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# 1  
Old 12-26-2012
find and replace with variable -sed
Hi,

I have a ksh script where I am trying to mask the password in the log files.

$loc - is my directory
$PGUIDE_DB_USER_PSW - is a variable that holds the password I am looking for

Code:
find $loc/logs -type f -exec sed -i "s/$PGUIDE_DB_USER_PSW/*****/"g {} \;

I get an error:

ksh: /*****/g: not found
sed: illegal option -- i


Any ideas?
# 2  
Old 12-26-2012
I am not a expert but i making a guess , the sed version you are using does not have -i option in it. I believe you are using a solaris machine.
# 3  
Old 12-26-2012
try also:
Code:
find $loc/logs -type f -exec sed -i 's?$PGUIDE_DB_USER_PSW?*****?g' {} \;

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