Bash function: extraxt different fields from JSON with grep regex without using jq

Question

How can I write a bash function to extract a user defined field from JSON?

#!/bin/bash
JSON='{"field1":"field1Value","field2":"field2Value"}'
function getFieldValueFromJson {
    JSON_FIELD=$1
    JSON=$2
    regex=".\"$JSON_FIELD\":\".*\""
    key=$(echo $JSON |grep -Po $regex)
    echo $key
}
getFieldValueFromJson field1 $JSON
getFieldValueFromJson field2 $JSON

Output is:

{"field1":"field1Value","field2":"field2Value"

,"field2":"field2Value"

Any idea how to solve this problem, if JSON is never pretty printed?

Edit:

OK, I can use jq but I want to solve this just with grep and regex.

There are multiple issues possible. Do check http://shellcheck.net — anishsane, Aug 02 '16 at 10:00
In [Merge two json in bash (no jq)](http://stackoverflow.com/q/38659763/) you can read some interesting comments on why it is not a good idea to try to parse JSON with Bash instead of using the tool for it (jq). — fedorqui, Aug 02 '16 at 10:17

score 1 · Accepted Answer · answered Aug 02 '16 at 13:02

1

With regex="\"$JSON_FIELD\":\"[^(\",)]*\""

O/P:

"field1":"field1Value"
"field2":"field2Value"

answered Aug 02 '16 at 13:02

jijinp

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Bash function: extraxt different fields from JSON with grep regex without using jq

1 Answers1