How to extract version number from an xml attribute with bash - regex

Question

From this xml string from my config.xml file I need to extract the first three digits of the version number:
<widget id="com.test.enterprise.test" version="3.0.0.0" xmlns="http://www.w3.org/ns/widgets" xmlns:cdv="http://cordova.apache.org/ns/1.0">
I've tried:
cat config.xml | grep "<widget" | sed 's/[^0-9.]*\([0-9.]*\).*/\1/'
but this only yields a . How would the correct regex look like?

Answer 1

Don't use regular expressions to parse XML.
xmllint -xpath 'string(//*[local-name()="widget"]/#version)' 1.xml \
| cut -f1-3 -d.
If you need to specify the namespace, too, use the namespace-uri function:
//*[local-name()="widget"][namespace-uri()="http://www.w3.org/ns/widgets"]

Answer 2

GNU grep with PCRE support \K don't include left of '\K' in the result
grep -Po '<widget.*?version="\K[^"]*' <<< '<widget id="com.test.enterprise.test" version="3.0.0.0" xmlns="http://www.w3.org/ns/widgets" xmlns:cdv="http://cordova.apache.org/ns/1.0">'
To have only first 3 digits
grep -Po '<widget.*?version="\K\d*(\.\d*){2}' <<< '<widget id="com.test.enterprise.test" version="3.0.0.0" xmlns="http://www.w3.org/ns/widgets" xmlns:cdv="http://cordova.apache.org/ns/1.0">'

Answer 3

You may grab the digits and dots only after version=" substring:
cat config.xml | grep "<widget" | sed 's/.*version="\([0-9.]*\).*/\1/'
See the online demo
Pattern details:
.* - any 0+ chars
version=" - a version=" substring
\([0-9.]*\) - capturing group #1 matching zero or more digits or .
.* - any 0+ chars.
The \1 backreference will keep Group 1 value in the result.

Answer 4

For first three digits of version:
grep -oP 'widget.*version="\K\d+\.\d+\.\d+' xmlFile
3.0.0

Answer 5

try following awks too, hope this may help you too.
solution 1st: Using match function of awk.
awk '{match($0,/version=\"[^"]*/);print substr($0,RSTART+9,RLENGTH-9)}' Input_file
solution 2nd: Going through one by one all the fields and then checking for version in them.
awk '{for(i=1;i<=NF;i++){if($i ~ /version/){gsub(/version=|\"/,"",$i);print $i;next}}}' Input_file
solution 3rd: making record separator as space and field separator as (").
awk -v RS=" " -v FS="\"" '/^version/{print $2}' Input_file
solution 4th: simply substituting all the text from starting to till string version=" then again substituting from " to till end, which will keep only version number in output.
awk '{sub(/.*version=\"/,"");sub(/\".*/,"");print}' Input_file
I hope this helps.