b= find =/home/root/dir1/*.xml | xargs grep '</XML>' | cut -d ":" -f1 | uniq

From a folder that has many files, I am trying to get the filename which has </XML> in it using the above command. But while giving the result, the variable b is showing the path also.

Ex: b=/homt/root/dir1/gg.xml.
How can I just get the value gg.xml in b?
The path I am giving as a variable.It is not hardcoded as (/home/root/dir1/*.xml)
Ex: $path/*.xml.So the occurrence of "/" can vary.

I am using KSH.

Thanks,
Floyd

One options:

b=$(find $MY_PATH -exec grep -l '</XML>' {} \; | xargs -I{} basename {} | uniq -u)

Hi Scott,
It does not print anything for me.It just come out of the shell. My filenames are like file1.xml,file2.xml,...
Also I tried echo ${b##*/}. That also does nothing.Bad luck..

One way ..

$ b=/homt/root/dir1/gg.xml
$ echo $b
/homt/root/dir1/gg.xml
$ echo $(basename $b)
gg.xml

You said that the path you wanted to search was in a variable. Did you change $MY_PATH to something meaningful for you?

I did test this before I posted it!

$ cat file1.xml
<XML>
this is some xml
</XML>

$ cat file2.xml
<XML>
this is some more xml
</XML>

$ MY_PATH=.

$ b=$(find $MY_PATH -exec grep -l '</XML>' {} \; | xargs -I{} basename {} | uniq -u>

# echo $b
file1.xml file2.xml

Thanks friends