sed regexp in a bash script

Question

Hello!

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!

Answer 1

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.

Answer 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.

Answer 3

$ 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