Inserting String variable into JSON which is in a form of String

Question

I have a JSON payload saved as a String

String jsonBody = “{\n”
            + ” \“example\“: {\n”
            + ”  \“example\“: [\n”
            + ”   {\n”
            + ”    \“example\“: 100,\n”
            + ”    \“this_is_example_json_key\“: \“this_is_example_json_value\“,\n”

I created that by copying body from i.e Postman into

String jsonBody = "here I pasted the body";

Unfortunately I cannot have everything hardcoded there, so I have to change some values to variables. The JSON in postman looks like:

"this_is_example_json_key":"x"

And so on. Let's assume that:

String x = “this_is_example_json_value“;

If I just replace it like

+ ” \“this_is_example_json_key\“: \“ + x + \“,\n”

or something like that, the value in the body will be just this_is_example_json_value, where I need "this_is_example_json_value" (the "" marks are part of the value).

So the question is, how to set up those + / " in the String, so in the end in the value of the JSON I will end up with the value inside " ".

I've tried to play with the " / + but nothing of those were working. Variable must be passed with those " " because otherwise, the API is sending back an error.

It's unclear what JSON you want to end up with. Please add that to your question. — tgdavies
– tgdavies, Commented Dec 7, 2022 at 3:20

Antonio Berritto · Accepted Answer · 2022-12-13 21:35:12Z

2

Since java 15, if you want only use the string, you can also do in this way:

int this_is_example_json_value= 100;
String json = """
{
    "this_is_example_json_key":  %d
}
""".formatted(this_is_example_json_value);

Here the official jep.

answered Dec 13, 2022 at 21:35

Antonio Berritto

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Comments

tgdavies · Accepted Answer · 2022-12-07 03:22:33Z

2

Don't try to build up JSON using strings. Use a proper JSON parser.

import org.json.JSONException;
import org.json.JSONObject;

public class Eg {
    public static void main(String[] args) throws JSONException {
        String x = "this_is_example_json_value";
        JSONObject example = new JSONObject();
        example.put("this_is_example_json_key", x);
        System.out.println(example.toString());
    }
}

Which outputs:

{"this_is_example_json_key":"this_is_example_json_value"}

With no messing around wondering what needs to be escaped.

edited Dec 7, 2022 at 3:22

answered Dec 7, 2022 at 2:39

tgdavies

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1 Comment

Adam Over a year ago

I will keep that in mind when I will be able to, but in this case I have there, there is just no point of doing that. But really appreciate your help, thank you!

Serge · Accepted Answer · 2022-12-07 03:43:04Z

1

you can use an extra " \ " "

    String x = "this_is_example_json_value";

    String jsonBody = "{\n"
            + "\"example\": {\n"
            + "  \"example\": [\n"
            + "  {\n"
            + " \"example\": 100,\n"
            + "\"this_is_example_json_key\":" + "\"" + x + "\"" + "\n }"
            +"\n  ]\n   }\n    }";

in this case you will get a json string

{
  "example": {
    "example": [
      {
        "example": 100,
        "this_is_example_json_key": "this_is_example_json_value"
      }
    ]
  }
}

edited Dec 7, 2022 at 3:43

answered Dec 7, 2022 at 2:17

Serge

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2 Comments

Serge Over a year ago

@tgdavies I am sorry , but the json string is shown already, why it should be repeated?

tgdavies Over a year ago

Sorry I meant to put that comment on the Question!

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