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Parsing xml files

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# 1  
Old 09-26-2013
Parsing xml files
I want to search for all the xml files on the server that have "Status" in them. Is this the correct code that I should use? Can anyone explain exactually what this code does?
Code:
xmlFileNames=$(find . -name "*.xml" -exec grep -l ".*Status" {} \; 2>/dev/null)

# 2  
Old 09-26-2013
It searches for all XML files in the current directory hierarchy (find . -name "*.xml") for the string Status (-exec grep -l ".*Status") (the .* in this search pattern is superfluous), returning the names of the files (grep -l) that do into a variable called xmlFileNames (xmlFileNames=$(...)).
This User Gave Thanks to Scott For This Post:
emc^24sho
# 3  
Old 09-26-2013
Thanks Scott!

---------- Post updated at 10:57 AM ---------- Previous update was at 10:56 AM ----------

Is my code correct then? lol
# 4  
Old 09-26-2013
Only if you run it from / (the root directory), otherwise:

Code:
xmlFileNames=$(find / ...)

would be required (/ instead of .)

You might encounter a lot of errors (access denied, or other) along the way, so:
Code:
xmlFileNames=$(find / ... 2> /dev/null)

(put 2> /dev/null inside the closing parenthesis to throw away the errors)
This User Gave Thanks to Scott For This Post:
emc^24sho
# 5  
Old 09-26-2013
Thanks a bunch Scott!
# 6  
Old 09-26-2013
".*Status" is a bit redundant, "Status" would do.
This User Gave Thanks to Corona688 For This Post:
emc^24sho
# 7  
Old 09-27-2013
If I use a
Code:
 xmlfilename=$(grep -l "Status" *.xml )

can I pull out each file name seperately in a for loop?
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