Shell script to find weblogic home directory


 
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# 1  
Old 09-11-2014
Shell script to find weblogic home directory

Hi,

I am trying to find the weblogic home directory whether it is installed in the Linux box. if it is existing display the weblogic home and the corresponding Java home directory else display message as that wl home does not exist. I already wrote a program but it was not working properly.
Please help me

Code:
#!/bin/ksh
WL_LOCATION=`find / -name wlserver_*`
if [-d $WL_LOCATION ] then
echo "WL_HOME ......... $WL_LOCATION"
cd $WL_LOCATION/common/bin
echo `pwd`
echo `sed s/\"//g grep JAVA_HOME commEnv.sh | grep jdk`
else echo "WL Home does not exists ...."
fi

Regards,
Spgreddy

---------- Post updated at 02:07 AM ---------- Previous update was at 12:21 AM ----------

Hi

I modified the script as shown below.

Code:
#!/bin/ksh
WL_LOCATION=`find / -name wlserver_*`
echo " $WL_LOCATION"
if [! -d $WL_LOCATION]; then
        echo "WL HOME does not exists ..." exit 1
        exit 1
else {
        echo "WL_HOME ......... $WL_LOCATION"
        cd $WL_LOCATION/common/bin
        echo `grep JAVA_HOME commEnv.sh | grep jdk|sed 's/"//g'`
      }
fi

when I execute, the below is the output.

Code:
/u01/app/oracle/Middleware/wlserver_10.3
./test1.sh[5]: [!: not found [No such file or directory]
WL_HOME ......... /u01/app/oracle/Middleware/wlserver_10.3
JAVA_HOME=/u01/app/jdk1.6.0_02


But here I am not able to understand that why it shows like this even it provides the actual output.
Code:
./test1.sh[5]: [!: not found [No such file or directory]


Please help us how to over come this?

Last edited by radoulov; 09-12-2014 at 05:44 AM.. Reason: Add CODE tags.
# 2  
Old 09-11-2014
There are several things that look strange in your script and I'm sure it isn't doing what you are trying to do. There are missing quotes, missing seicolons, extraneous echoes, unnecessary command substitutions, and I have no idea what the sed command is trying to do. But, since I don't know where weblogic might be installed on your server, I can't guess at what you are trying to do.

Is there a user named weblogic on your server?

After looking at every file on your system, are you sure that there will be exactly one file with a name starting with the characters wlserver_? Are there any constraints on where that one file could be located?

Please describe in English the logic behind each step in your script so we can help you modify your code to fix any problems you are running into.
# 3  
Old 09-12-2014
Hi Don,

Thank you for your support. Forget above script currently I am looking a shell script by reading the string (/u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar) and display required output as shown below

/u01/app/oracle/Middleware/wlserver_10.3

Please help us to resolve this

Regards,
Spgreddy
# 4  
Old 09-12-2014
Quote:
Originally Posted by spgreddy
Hi Don,

Thank you for your support. Forget above script currently I am looking a shell script by reading the string (/u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar) and display required output as shown below

/u01/app/oracle/Middleware/wlserver_10.3

Please help us to resolve this

Regards,
Spgreddy
I can't figure out what this means. Does the string you are reading contain parentheses as shown? Is this string in a file? Is this string in a shell variable?

How is your script supposed to know what part of the string is the part you want to return? Are you trying to remove the last three components of the pathname stored in the string? Are you trying to remove all components of the pathname after the component that starts with wlserver? Are you trying to remove all components from the pathname stored in the string after the component that ends with a decimal digit? Are you trying to remove the 6th / and everything that follows it from your string?

What shell are you using? What OS are you using?
# 5  
Old 09-12-2014
Hi DOn
The string contains /u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar

For example :
Code:
x=/u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar

i am trying to remove the last three componets /server/lib/weblogic.jar
and want to display the output as /u01/app/oracle/Middleware/wlserver_10.3

currently i am using bash sheel in redhat linux 5

Regards,
Spgreddy

Last edited by Don Cragun; 09-12-2014 at 02:21 AM.. Reason: Add CODE and ICODE tags.
# 6  
Old 09-12-2014
So try something like:
Code:
#!/bin/bash
x=/u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar
y=${x%/*/*/*}
printf 'x is %s\ny is %s\n' "$x" "$y"

which produces:
Code:
x is /u01/app/oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar
y is /u01/app/oracle/Middleware/wlserver_10.3

This User Gave Thanks to Don Cragun For This Post:
# 7  
Old 09-12-2014
Hi Don,

I have one more issue. A variable contains value like this

Code:
jhome=JAVA_HOME=/u01/app/jdk1.6.0_02

now I want output as
Code:
jhome=/u01/app/jdk1.6.0_02

please help me how can I write shell script for this

Regards,
Spgreddy

Last edited by radoulov; 09-12-2014 at 05:44 AM..
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