[SOLVED] bash + how to print sequence lines in the same line

Issue

I created the following find syntax in order to search all files that start with moon_ and ended with .xml and should contain the tag - <Version>

here is example

find   /main_folder/kits/  -maxdepth 2  -name "moon*" -name "*.xml" -exec grep '<Version>' {} \;    -exec echo {} \;
 
    <Version>1.3.12-dev.137</Version>
/main_folder/kits/A/moon_aaa.xml
    <Version>2.1.1-dev.13</Version>
/main_folder/kits/B/moon_bbb.xml
        <Version>1.0.144</Version>
/main_folder/kits/C/moon_ccc.xml

as above its print the path and the version

but we want to print is better so path and version will be in the same line like the following example

find   /main_folder/kits/  -maxdepth 2  -name "moon*" -name "*.xml" -exec grep '<Version>' {} \;    -exec echo {} \; | .........
 
    
/main_folder/kits/A/moon_aaa.xml <Version>1.3.12-dev.137</Version>
/main_folder/kits/B/moon_bbb.xml <Version>2.1.1-dev.13</Version>
/main_folder/kits/C/moon_ccc.xml <Version>1.0.144</Version>

what we need to add in our syntax in order to print both in the same line ?

Solution

Like this:

find /main_folder/kits/ -maxdepth 2 -name 'moon*' -name '.xml' -exec bash -c '
    for xml; do 
        res=$(grep "<Version>" "$xml") && echo "$1 $res"
    done
' bash {} +

Or better, instead of grep, use :

xmllint --xpath '//Version/text()' "$xml"

Answered By - Gilles Quénot

Answer Checked By - Mary Flores (WPSolving Volunteer)