Shell Programming and Scripting

i have a text file consists of different file names like:

line 1: lib/libIMb.so message broker file
line 2: lil/imbdfg.lil message broker file

i need to extract libIMb.so and imbdfg.lil files from those lines and save them in a variable. so that i can search for those files in different directories.

Use Parameter Substitution in Korn shell or POSIX shell:

Code:

#!/bin/ksh

while read line
do
        file=${line#*/}
        file=${file%% *}
        echo $file
done < filename

Or use sed

Code:

sed 's/.*\///; s/ .*//' filename

You could try this:

Code:

set -- $(awk 'NR<3 { sub(".*/","",$1); print $1 }' names.txt)

$1 is file1 $2 is file2

Thank you for the replies. But i am not getting the required output when i try these. May be i am not clear. sorry for that.
I have a text file readme.txt, which consists of all these filenames in between the lines.

lib/libImbCmdLib.so Message File
lib/libImbRdl.so Message File
lib/libJDBCType4JNISinglePhaseTrxn.so Message File
lib/libJMSJNILocalTrxn.so Message File
lib/libMQLibrary.so Message File

from the above lines i need to extract the string with the filename like libImbCmdLib.so
libImbRdl.so
libJDBCType4JNISinglePhaseTrxn.so and save in a variable.

By reading that variable i need to find this filename in a directory1 and if match found need to copy that in directory2. This has to repeat for all the filenames in the textfile. Any logic and script will be appreciated. Thanks.

Code:

sed -e 's:^.*/::' -e 's/ .*//' readme.txt | while read file
do
    find directory1 -name $file -exec cp {} directory2/
done

Thanks for the reply. But i m getting error:

find: missing argument to '-exec'
find: missing argument to '-exec'
find: missing argument to '-exec'
find: missing argument to '-exec'

Sorry missed \; on end of find statement:

Code:

sed -e 's:^.*/::' -e 's/ .*//' readme.txt | while read file
do
    find directory1 -name $file -exec cp {} directory2/ \;
done