Extracting JSON variable using bash

Question

I need to extract the variable from a JSON encoded file and assign it to a variable in Bash.

excerpt...from file.json

  "VariableA": "VariableA data", 
    "VariableB": [
        "VariableB1", 
        "VariableB2", 
        "VariableB3", 
        "VariableB3"
    ],

I've gotten somewhere with this

variableA=$(fgrep -m 1 "VariableA" file.json )

but it returns the whole line. I just want the data

For the VariableB I need to replace the list with comma separated values.

I've looked at awk, sed, grep, regexpressions and really given the learning curve...need to know which one to use, or a better solution.

Thanks for your suggestions...but this is perfect git://github.com/kristopolous/TickTick.git

Dennis · Accepted Answer · 2012-08-11 23:01:35Z

7

You are better off using a JSON parser. There are many listed at http://json.org/ including two for the BASH shell.

answered Aug 11, 2012 at 23:01

Dennis

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1 Comment

Clayton Stanley Over a year ago

jsawk worked for me. And it was fairly easy to install on OS X.

tworec · Accepted Answer · 2017-07-13 15:10:20Z

4

There is powerful command-line JSON tool jq.

Extracting single value is easy:

variableA=$(jq .VariableA file.json)

For comma separated array contents try this

variableB=$(jq '.VariableB | @csv' file.json)

or

variableB=$(jq '.VariableB | .[]' file.json | tr '\n' ',' | head -c-1)

edited Jul 13, 2017 at 15:10

answered Jul 13, 2017 at 15:03

tworec

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Comments

szabgab · Accepted Answer · 2014-04-28 08:35:05Z

0

If you're open to using Perl they have a 'open()' function that will pipe a file with the json function 'to_json'. And if you want to extract json you can use the 'from_json' function. You can check it out here:

http://search.cpan.org/~rjbs/perl-5.16.0/lib/open.pm

http://metacpan.org/pod/JSON#to_json

http://metacpan.org/pod/JSON#from_json ( you might also try using decode json as well)

edited Apr 28, 2014 at 8:35

szabgab

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answered Aug 11, 2012 at 23:17

John Verber

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