sed regexp in a bash script

Question

I want to extract a certain part of a string, if it exists. I'm interested in the xml filename, i.e i want whats between an "_" and ".xml".

This is ok, it prints "555"

MYSTRING=`echo "/sdd/ee/publ/xmlfile_555.xml" | sed 's/^.*_\([0-9]*\).xml/\1/'`
echo "STRING = $MYSTRING"

This is not ok because it returns the whole string. In this case I don't want any result. It prints "/sdd/ee/publ/xmlfile.xml"

MYSTRING=`echo "/sdd/ee/publ/xmlfile.xml" | sed 's/^.*_\([0-9]*\).xml/\1/'`
echo "STRING = $MYSTRING"

Any ideas how to get an "empty" result in the second case.

thanks!

David Z · Accepted Answer · 2010-06-29 18:04:26Z

2

You just need to tell sed to keep its mouth shut if it doesn't find a match. The -n option is used for that.

MYSTRING=`echo "/sdd/ee/publ/xmlfile_555.xml" | sed -n 's/^.*_\([0-9]*\)\.xml/\1/p'`

I only made two changes to what you had: the aforementioned -n option to sed, and the p flag that comes after the s/// command, which tells sed to print the output only if the substitution was successfully done.

EDIT: I've also escaped the final . as suggested in the comments.

edited Jun 29, 2010 at 18:04

answered Jun 29, 2010 at 11:53

David Z

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2 Comments

Douglas Leeder Over a year ago

You probably also want to escape the final .

David Z Over a year ago

@Douglas: true, I hadn't noticed that (and that's why you don't copy and paste code ;-)

Noufal Ibrahim · Accepted Answer · 2010-06-29 11:44:52Z

0

Try this?

basename /sdd/ee/publ/xmlfile_555.xml | awk -F_ '{print $2}'

The output is 555.xml

With the other one.

basename /sdd/ee/publ/xmlfile.xml | awk -F_ '{print $2}'

The output is an empty string.

answered Jun 29, 2010 at 11:44

Noufal Ibrahim

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1 Comment

Noufal Ibrahim Over a year ago

It wasn't meant to be an "out of the box solution" as much as a possible approach but yeah, your're right.

ghostdog74 · Accepted Answer · 2010-06-29 12:54:22Z

0

$ path=/sdd/ee/publ/xmlfile_555.xml
$ echo ${path##*/}
xmlfile_555.xml
$ path=${path##*/}
$ echo ${path%.xml}
xmlfile_555
$ path=${path%.xml}
$ echo ${path##*_}
555

answered Jun 29, 2010 at 12:54

ghostdog74

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