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Generate class path dynamically based on source path

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Generate class path dynamically based on source path

Hi experts,

I have multiple file names ending with .jsp located in
$SOME_DIR, $SOME_DIR/f1/,$SOME_DIR/f2/test,$SOME_DIR/f3/fa

and there are equivalent class files in
 $SOME_DIR/WEB-INF/classes/_pages,$SOME_DIR/WEB-INF/classes/_pages/_f1, $SOME_DIR/WEB-INF/classes/_pages/_f2/_test,$SOME_DIR/WEB-INF/classes/_pages/_f3/_fa

I have to write a script to match class file for each jsp file
Name: test.jsp 
Located in $SOME_DIR/f2/test
I have to check for _test.class in $SOME_DIR/WEB-INF/classes/_pages/_f2/_test (There is underscore in the path of the class)

if this class is not present then do something

I am doing find on $SOME_DIR for jsps and running basename on

for i in $( find $SOME_DIR -name '*.jsp'|  sed "s|$SOME_DIR/||;"); 
echo jsp file $i
class_name==_$(echo $(basename $i)|cut -d'.' -f1).class
echo class name is $class_name
#now this _test.class file is present  in $SOME_DIR/WEB-INF/classes/_pages/_f2/_test
#$SOME_DIR/WEB-INF/classes/_pages directory or its sub directories is where all class files are present

Source of jsp is $SOME_DIR/f2/test
Destination of equivalent class is located in #$SOME_DIR/WEB-INF/classes/_pages/_f2/_test

logical code of destination path for a given .jsp will be $SOME_Dir+WEB_INF+classes+_pages+(dirname $i minus $SOME_DIR with underscore)

I am not really sure how to generate class path based on source path dynamically.

---------- Post updated 12-19-16 at 04:27 PM ---------- Previous update was 12-18-16 at 05:25 PM ----------

Resolved myself. I will post the solution...

Last edited by oraclermanpt; 12-18-2016 at 07:05 PM.. Reason: fixed spell mistakes

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