awk or grep to get a specific string in shell

manpreet Best Answer 2 years ago

I am trying to get only the value of percentage part from this data which is stored in a variable also I want to get only the first percentage value that is present in the output.

 id="09EhL_map" name="09EhL_map">
 shape="poly" coords="374,274,374,274,376,274,376,274,376,274,376,274" title="5-April-2019: 0.0% of runs for TT1 were successful" alt="" nohref="nohref"/>
 shape="poly" coords="368,274,368,274,371,274,374,274,374,274,374,274" title="4-April-2019: 20.0% of runs for TT2 were successful" alt="" nohref="nohref"/>

what I have done

var1="_map" name="09EhL_map">
poly" coords="374,274,374,274,376,274,376,274,376,274,376,274" title="5-April-2019: 0.0% of runs for HIP-HCTP-HIP3 were successful" alt="" nohref="nohref"/>
poly" coords="368,274,368,274,371,274,374,274,374,274,374,274" title="4-April-2019: 0.0% of runs for HIP-HCTP-HIP3 were successful" alt="" nohref="nohref"/>"

var2=$(echo var1 | grep "%")

This doesn't return anything.

manpreet 2 years ago

You can use:

var2=$(echo $var1 | grep -o  [0-9][0-9]*\.[0-9][0-9]*%)

here grep -o [0-9][0-9]*\.[0-9][0-9]*% will match floating point numbers having % sign in the end.

In your case it will print

0.0%
20.0%

If you want to get only the first match, then use:

var2=$(echo $var1 | grep -o  [0-9][0-9]*\.[0-9][0-9]*% | head -n1)