Reliable JSON string validator in Java

Question

I need some help coming with a reliable JSON string validator - a method which intake a string and checks if it's a valid JSON. Example: if I pass {"color":"red"} or {"amount":15} it will pass but something like "My invalid json" will not. In short I need something that is as reliable as www.jsonlint.com validator. BTW - I'm not interested in deserializing into java object, because that's not my requirement. I may receive an arbitrary string and all I have to do is validate it has a valid JSON format.

I have already researched on this forum several posts on the subject regarding java JSON string validations.

What I have done so far:

I tried using these classes: org.json.JSONObject and org.json.JSONArray in the following manner:

private static boolean isValidJSONStringObject(String requestBody){
    try {
        new JSONObject(requestBody);
    } catch (JSONException jsonEx) {
        return false;
    }
    return true;
}

private static boolean isValidJSONStringArray(String requestBody) {
    try {
        new JSONArray(requestBody);
    } catch (JSONException jsonEx) {
        return false;
    }
    return true;
}

However, the following strings (entire lines) still go through, and they shouldn't:

{"color":"red"}{"var":"value"}

[1,2,3][true,false]

in other words when I have objects/arrays repeated w/out any encapsulation in some parent object. If you paste these lines in www.jsonlint.com validator they both fail.

I know there is always a regex option but I gess that cannot be guaranteed 100% because of the recursive nature of JSON and those regex expressions are going to be rather complex.

Any help will be greatly appreciated!

Did you try other libraries? I think jackson doesn't allow that, for instance. — yshavit, Apr 03 '13 at 15:46
`"My invalid json"` is a bad example of invalid JSON, because it is, in fact, a valid JSON string. — pfyod, Apr 03 '13 at 15:54

Perception · Answer 1 · 2013-04-03T18:10:24.580

6

Gson can handle this. Here is an example:

public boolean isValid(String json) {
    try {
        new JsonParser().parse(json);
        return true;
    } catch (JsonSyntaxException jse) {
        return false;
    }
}

String json = "{\"color\":\"red\"}{\"var\":\"value\"}";
System.out.println(isValid(json));

Note that Gson does allow some leniency in input JSON, which might not be desired. For example, unquoted keys are automatically converted to quoted ones by the parser. This may or may not be a deal-breaker depending on your expected usage.

edited Apr 03 '13 at 18:10

answered Apr 03 '13 at 15:52

Perception

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1

The Jackson library doesn't seem to handle this unless you know what type of object you're going to deserialize to. This Gson library is better, but seems a bit lenient. For example, it converts `abc` to `"abc"` instead of throwing an error. But if you limit what's acceptable to just `.isJsonObject()` (method of `JsonElement`) then it works well. – Allan Apr 03 '13 at 17:43
1

@Allan - Jackson ***can*** parse JSON without an end object type (by using one of the `readTree` methods for example). But it does not (by default) report back reliably for invalid JSON. Gson leniency mode is soft in some regards, and latest supposed bug fixes have evidently not held. Thanks for the feedback, I've updated my answer. – Perception Apr 03 '13 at 18:08
Yes, given the "soft leniency mode" of gson parser we decided (for now) to complement with jackson parser to satisfy all of our requirements. – rich Apr 03 '13 at 20:49

score 3 · Answer 2 · answered Apr 04 '13 at 16:58

Here is our solution for now. Using the two different libraries (gson - first private method and jackson - second private method) is not ideal but at least we are passing all of our unit/integration tests. I bet we can do all we need with just the jackson tools.

public static boolean isStringValidJSON(String jsonString) {
    return (isJSONStringObjectOrArray(jsonString) && isJSONStringParsable(jsonString));
}

private static boolean isJSONStringObjectOrArray(String jsonString) {
    try {
        JsonElement element = new JsonParser().parse(jsonString);

        return (element.isJsonObject() || element.isJsonArray());
    } catch (JsonSyntaxException jsonEx) {
        return false;
    }
}

private static boolean isJSONStringParsable(String jsonString) {
    try {
        org.codehaus.jackson.JsonParser parser = 
          new ObjectMapper().getJsonFactory().createJsonParser(jsonString);
        while(parser.nextToken() != null) {
        }
        return true;
    } catch (JsonParseException e) {
        return false;
    } catch (IOException e) {
        return false;
    }
}

score -3 · Answer 3 · edited Jan 29 '16 at 10:34

-3

Paste your string here. See the output.

EDIT:

The link above does not work anymore, this is a good alternative.

edited Jan 29 '16 at 10:34

Pascal Schneider

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answered Jun 20 '13 at 18:26

Prashant

3
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3

the OP is trying to do this programatically. not by a tool. – SMR Jan 29 '16 at 10:34

Reliable JSON string validator in Java

3 Answers3

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