Issue
I'd like to extract the "id"
key from this single line of JSON.
I believe this can be accomplished with grep, but I am not sure on the correct way.
If there is a better way that does not have dependencies, I would be interested.
Here is my example output:
{"data": {"name": "test", "id": "4dCYd4W9i6gHQHvd", "domains": ["www.test.domain.com", "test.domain.com"], "serverid": "bbBdbbHF8PajW221", "ssl": null, "runtime": "php5.6", "sysuserid": "4gm4K3lUerbSPfxz", "datecreated": 1474597357}, "actionid": "WXVAAHQDCSILMYTV"}
Solution
If you have a grep that can do Perl compatible regular expressions (PCRE):
$ grep -Po '"id": *\K"[^"]*"' infile.json
"4dCYd4W9i6gHQHvd"
-P
enables PCRE-o
retains nothing but the match"id": *
matches"id"
and an arbitrary amount of spaces\K
throws away everything to its left ("variable size positive look-behind")"[^"]*"
matches two quotes and all the non-quotes between them
If your grep can't do that, you an use
$ grep -o '"id": *"[^"]*"' infile.json | grep -o '"[^"]*"$'
"4dCYd4W9i6gHQHvd"
This uses grep twice. The result of the first command is "id": "4dCYd4W9i6gHQHvd"
; the second command removes everything but a pair of quotes and the non-quotes between them, anchored at the end of the string ($
).
But, as pointed out, you shouldn't use grep for this, but a tool that can parse JSON – for example jq:
$ jq '.data.id' infile.json
"4dCYd4W9i6gHQHvd"
This is just a simple filter for the id
key in the data
object. To get rid of the double quotes, you can use the -r
("raw output") option:
$ jq -r '.data.id' infile.json
4dCYd4W9i6gHQHvd
jq can also neatly pretty print your JSON:
$ jq . infile.json
{
"data": {
"name": "test",
"id": "4dCYd4W9i6gHQHvd",
"domains": [
"www.test.domain.com",
"test.domain.com"
],
"serverid": "bbBdbbHF8PajW221",
"ssl": null,
"runtime": "php5.6",
"sysuserid": "4gm4K3lUerbSPfxz",
"datecreated": 1474597357
},
"actionid": "WXVAAHQDCSILMYTV"
}
Answered By - Benjamin W. Answer Checked By - Senaida (WPSolving Volunteer)