UNIX for Dummies Questions & Answers

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:

Code:

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

puts all the filenames which match into b.

Not so great if any file names have spaces in them.

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 ..

Code:

$ 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!

Code:

$ 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