UNIX for Advanced & Expert Users

Hi All,

I have the file in the following format.I need to change the password "tomcat","admin","mgr" and "testing" in the file with the encrypted passwords.
The encrypted passwords are given to me by another script.
<?xml version='1.0' encoding='utf-8'?>
<tomcat-users>
<role rolename="tomcat"/>
<role rolename="role1"/>
<role rolename="manager"/>
<role rolename="admin"/>
<user username="tomcat" password="tomcat" roles="tomcat"/>
<user username="role1" password="admin" roles="role1"/>
<user username="both" password="mgr" roles="tomcat,role1"/>
<user username="admin" password="testing" roles="admin,manager"/>
</tomcat-users>
Following is the script that I had started, But I am stuck with the perl -i command.the perl -i command works like a charm with the string, but not working with the variable. Can anyone please help.

Code:

#!/usr/bin/ksh
passwds=`awk -F"\"" '$3 ~ "password="{print $4}' tomcat-users.xml`
for i in $passwds
do
/opt/coreservices/tomcat-5.5.9/bin/digest.sh -a sha $i >temp
en_passwd=`awk -F ":" '{print $2}' temp`
perl -i.bak -pe's/(password=").*?"/$1$enpasswds"/' tomcat-users.xml

But this command works like a charm!!

Code:

perl -i.bak -pe's/(password=").*?"/$1abcd"/' tomcat-users.xml

Please let me know if sed -i command would work for this scenario.

Thanks for all the help!

Strange that $1 is substituted while enclose by single quotation marks, which prevent that for shell variables.

If you tend to use sed instead of perl (I am not that familiar with perl) you can just write something like (just an example):

Code:

sed 's/(password=").*?"/'${1}${enpasswds}'"/' tomcat-users.xml  > newfile

$1 is not a shell variable but a Perl variable, since it is in the expression in single quotes passed to Perl with -e to execute, so there is no problem with it.
So is $enpasswds, but that has not been initialised in the Perl expression.

Thank you for all the replies!

The problem here is the sed command is not working.I need to make changes in the original file, tomcat-users file itself, so I tried the -i option. But that too isn't working..

Following id the command I tried.

Code:

sed -i's/(password=").*?"/'${1}${en_passwd}'"/'tomcat-users.xml

Can you please help!

As spirtle stated, $1 and $enpasswds are perl variables. In the upper part of your shell script you define en_passwds, but I see nowhere, that you feed perl with that ie. something like in awk when you use -v to handover a shell variable to become the value of an awk variable.

I don't know the -i switch for sed, sorry.

...

Code:

perl -i.bak -pe's/(password=").*?"/$1$ENV{en_passwd}"/' tomcat-users.xml

Do you want to use the same password for all users?

Thanks Zaxxon. I tried exporting the variable en_passwd, so that perl can recognize it. It recognized the variable , but was not able to change the file.

I am looking into other options(sed or awk) to achieve this. My biggest problem is I need to exactly replace only the password in the file, by a variable.

For eg: password="tomcat"

I need to replace only tomcat in quotes after the password word, no matter how many times tomcat word is int he file.

Will post the solution, if I am able to find it.

Thanks!
nua7