Invalid JSON Text - SQL (Java JSON Object)

Question

I'm putting in JSON into an SQL table (type: json) and it claims something is wrong - but I can't tell what it's pointing out directly.

Current json I'm trying to input:

'{"{\"type\":\"HUMAN\"}":"1.0"}'

Yes, it is "json within json" but the code I'm writing is kind of dependent on it being that way. I'm not willing to really change that part.

Here's the error:

#3140 - Invalid JSON text: "Missing a colon after a name of object member." at position 5 in value for column 'mines.materials'.

If you need code, I have it. I really need to know whats wrong with the JSON, if it requires a fundamental change to the code I can provide what I have.

That isn't valid JSON at all. If you pass it to JSON.parse in your browser, it throws. Use \\" not \". — Aluan Haddad
– Aluan Haddad, Commented Dec 9, 2017 at 3:29
that seemed to do the trick. JSONObject#toJSONString uses single back-slashes — Sage M
– Sage M, Commented Dec 9, 2017 at 3:34

Aluan Haddad · Accepted Answer · 2021-12-13 08:33:26Z

3

You should really try validating some basic assumptions.

'{"{\"type\":\"HUMAN\"}":"1.0"}' is not a valid JSON string as you can see below

"use strict";
(function () {
  try {
    JSON.parse('{"{\"type\":\"HUMAN\"}":"1.0"}');
  } catch (e) {
    console.error(e);
  }
}());

You are using attempting to use a JSON encoded string as a property key in another JSON encoded string but have not escaped it correctly.

You need to use \\" instead of \" to escape the quotes

"use strict";
(function() {
  try {
    console.log(JSON.parse('{"{\\"type\\":\\"HUMAN\\"}":"1.0"}'));
  } catch (e) {
    console.error(e);
  }
}());

edited Dec 13, 2021 at 8:33

answered Dec 9, 2017 at 3:36

Aluan Haddad

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