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Generating an xml having information related to files in the directory

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# 8  
Old 11-19-2009
Modify the selection pattern to meet your requirement :
Code:
ls -l  2>/dev/null | 
awk '
/\.xml$/ {

For example if you want to procced all Name.Version.<number>.xml files :
Code:
ls -l  2>/dev/null | 
awk '
/Name\.Version\.[0-9]+\.xml$/ {

Jean-Pierre.
# 9  
Old 11-19-2009
Great help...

But how did this code manage so many files..can you please explain a bit....

It will be really helpful..

---------- Post updated at 10:05 AM ---------- Previous update was at 09:50 AM ----------

In the line

/Name\.Version\.[0-9]+\.xml$/

can we include a variable defined in the script and passed in to awk

---------- Post updated at 10:31 AM ---------- Previous update was at 10:05 AM ----------

ls -l 2>/dev/null |
awk -v var=$mydate '
/Name\.Version\.[0-9]+\.xml$/ {

i want to ue "var" in the search pattern
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