Get JSON value from URL in BASH

Question

I need to get the "snapshot" value at the top of the file from this url: https://s3.amazonaws.com/Minecraft.Download/versions/versions.json

So I should get a variable that contains "14w08a" when I run the command to parse the json.

score 7 · Answer 1 · answered Feb 19 '14 at 18:38

7

This will do the trick

$ curl -s "$url" | grep -Pom 1 '"snapshot": "\K[^"]*'
14w08a

answered Feb 19 '14 at 18:38

Chris Seymour

3

+1 for this pragmatic approach (in general, @glennjackman's approach offers more flexibility); version for platforms where `-P` is not supported (e.g, OSX): `curl -s "$url" | egrep -m 1 '"snapshot"' | awk -F '"' '{ print $4 }'` – mklement0 Feb 19 '14 at 22:12
`-Pom` has three options: [`-P` is for `--perl-regexp`, and `-o` is for `--only-matching`, and `-m` is for `--max-count`](https://man7.org/linux/man-pages/man1/grep.1.html). And p[`\K` is `perl-regexp` flavor for a variable-length lookbehind](https://stackoverflow.com/a/33573989/1175496), "`\K` tells the engine to pretend that the match attempt started at this position."; another [example shown here](https://stackoverflow.com/a/5081519/1175496) – Nate Anderson Dec 28 '22 at 14:02

score 4 · Accepted Answer · edited Feb 19 '14 at 22:00

4

best thing to do is use a tool with a JSON parser. For example:

value=$(
  curl -s "$url" |
  ruby -rjson -e 'data = JSON.parse(STDIN.read); puts data["latest"]["snapshot"]'
)

edited Feb 19 '14 at 22:00

mklement0

answered Feb 19 '14 at 18:26

glenn jackman

score 0 · Answer 3 · answered Aug 25 '23 at 20:13

0

$ curl -s "$url" | jq -r .snapshot
14w08a

answered Aug 25 '23 at 20:13

notacorn

3 Answers3