How to extract a string after matching characters from a variable in shell script

Question

I have a file with following text as below

classA = Something
classB = AB1234567
classC = Something more
classD = Something Else

Objective:
Using a shell script, I want to read the text which says AB1234567 from above complete text.

So to start, I can read the second line of the above text using following logic in my shell script:

secondLine=`sed -n '2p' my_file`;
echo $secondLine;

secondLine outputs classB = AB1234567. How do I extract AB1234567 from classB = AB1234567 in my shell script?

Question:
Considering the fact that AB is common in that particular part of the text all the files I deal with, how can I make sed to read all the numbers after AB?

Please note that classB = AB1234567 could end with a space or a newline. And I need to get this into a variable

Not sure why my question was down-voted but anyways. Thanks a lot for the answers. — AdeleGoldberg, Oct 29 '20 at 16:36

αғsнιη · Accepted Answer · 2020-10-29T16:49:54.597

Try:

sed '2{ s/^classB = \(AB[^ ]*\) *$/\1/;q } ;d' your_fileName

2 is the line number.
- { open a sed group command.
  - s/ substitute below match
    - ^ is anchor for beginning of the line
    - $...$ is known a capture group with \1 as its back-reference
    - [^ ]* means any character but not a space
    - $AB[^ ]*$ capture AB followed by anything until first space seen but not spaces (back-reference is \1)
    - * means zero-or-more spaces
    - $ is anchor for end of the line
  - / with below
    - \1 back-reference of above capture group
  - / end of substitution
  - q quit to avoid reading rest of the file unnecessarily
- } close group command.
d delete any other lines before seen line number 2.

get into variable:

your_variableName=$(sed '2{ s/^classB = \(AB[^ ]*\) *$/\1/;q } ;d' your_fileName)

RavinderSingh13 · Answer 2 · 2020-10-29T16:24:19.263

3

Could you please try following, looks should be easy in awk. Considering you want to print 2nd line and print only digits in last field.

secondLine=$(awk 'FNR==2{sub(/[^0-9]*/,"",$NF);print $NF}' Input_file)

edited Oct 29 '20 at 16:24

answered Oct 29 '20 at 16:21

RavinderSingh13

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Thanks. Can you please use the variable `secondLine` and put this into another variable called `searchedtext` or something? – AdeleGoldberg Oct 29 '20 at 16:23
2

@AdeleGoldberg, Done, could you please let me know if this helped you? – RavinderSingh13 Oct 29 '20 at 16:24
2

This works as well. Confused which answer to choose. Such intelligent minds out there! Thanks a lot – AdeleGoldberg Oct 29 '20 at 16:34
1

@AdeleGoldberg, your welcome, its up to you which answer you select we all are here to learn, cheers and happy learning and sharing :) – RavinderSingh13 Oct 29 '20 at 16:34

score 2 · Answer 3 · answered Oct 29 '20 at 16:20

2

You may try this awk:

awk -F ' *= *' '$1 ~ /B$/ { print $2 }' file

AB1234567

answered Oct 29 '20 at 16:20

anubhava

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1

Thanks. Can you please use the variable `secondLine` and put this into another variable called `searchedtext` or something? – AdeleGoldberg Oct 29 '20 at 16:24
you can do `secondLine=$(awk -F ' *= *' '$1 ~ /B$/ { print $2 }' file)` to put output in a variable – anubhava Oct 29 '20 at 16:26

score 1 · Answer 4 · answered Oct 29 '20 at 16:25

I'm not 100% sure this is what you're looking for, but if you know there's only a single element in the file that starts with AB, this will get it into a variable:

$ cat sample.txt 
classA = Something
classB = AB1234567
classC = Something more
classD = Something Else
  
$ x=$(perl -ne 'print if s/^.*\s+(AB\S+)\s*$/$1/' sample.txt)
 
$ echo "the variable is: $x"
the variable is: AB1234567

Explanation of the regex:

^ beginning of line
.* anything
\s+ any number of spaces
(AB\S+) anything that starts with AB followed by non-spaces
\s*$ Zero or more spaces followed by the end of the line.

How to extract a string after matching characters from a variable in shell script

4 Answers4