How to read JSON in GET request query parameter

Question

I have to read JSON values from the URL using a query parameter for GET request. I am using Tomcat latest in a Spring Boot project.

       @RequestMapping(
            value = "/values",
            method = RequestMethod.GET, 
            headers = HttpHeaders.ACCEPT + "=" + MediaType.APPLICATION_JSON_VALUE,
            produces = "application/json")
       public ResponseEntity<String> myMethod(
            @RequestParam(value="key") String jsonRequestString) {
       //parse JSONString 
       //--
       }

GET request to the URL

Url:- http://localhost:port/values?key={"org":"AA","points":[{"name":"xy","Measures":"343"}]}]

Throws java.lang.IllegalArgumentException: Invalid character found in the request target. The valid characters are defined in RFC 7230 and RFC 3986

EDIT :- Sending JSON in original form leading me to nowhere, so working approach would be to send the JSON in an encoded form.

As the error says, that isn't a valid URL. You would need to URL-escape the JSON content before sending it, or (far better), use a POST and put the JSON in the request body. — Andrew McGuinness
– Andrew McGuinness, Commented Mar 20, 2017 at 11:57
I can't do any change in the request parameters or request type. — jones j alapat
– jones j alapat, Commented Mar 20, 2017 at 12:00
I need to somehow be able to accept those values.Any suggestion for that — jones j alapat
– jones j alapat, Commented Mar 20, 2017 at 12:02
This question maybe helpful: stackoverflow.com/questions/41053653/… In short you might not have an option but to downgrade to an older Tomcat version if the client is not escaping the query param. — Strelok
– Strelok, Commented Mar 20, 2017 at 12:30

ggorlen · Accepted Answer · 2022-05-17 01:51:17Z

2

You can encode the key parameter to Base64 format and append to the URL, then decode and use it in your controller method.

Key JSON text:

{"org":"AA","points":[{"name":"xy","Measures":"343"}]}]

Base64 encoded text:

eyJvcmciOiJBQSIsInBvaW50cyI6W3sibmFtZSI6Inh5IiwiTWVhc3VyZXMiOiIzNDMifV19XQ==

Decoded to again back to JSON text:

{"org":"AA","points":[{"name":"xy","Measures":"343"}]}]

edited May 17, 2022 at 1:51

ggorlen

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answered Mar 20, 2017 at 12:42

Aditya

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5 Comments

jones j alapat Over a year ago

I am not the one who is sending the request.I cannot change the request format .

Aditya Over a year ago

Check it out the solution proposed in this post stackoverflow.com/a/21689084/2633386

jones j alapat Over a year ago

still giving me Exception

Aditya Over a year ago

what exception are you getting? is that same or spring related ? in case any configuration issue check the comment in the link above

Aditya Over a year ago

Spring MVC 4, @RequestBody Map<String, Object> map works with the input json {"name":"frank","Type":"Lumber"}.

ProgrammerBoy · Accepted Answer · 2017-03-20 12:14:25Z

1

Few things you should take care of.

Use POST HTTP method to post your JSON to server.
Create a JAVA pojo class which should has a same structure as your JSON. Like for below JSON,
```
  {
        "id": 123,
        "status": "Ordered",
        "product": "Pepsi"
       } 
```

i would create a class( You can create a two separate class as you has array inside json) ,

public class Order {

private long id ;
private String status; 
private String product ;

public long getId() {
    return id;
}
public void setId(long id) {
    this.id = id;
}
public String getStatus() {
    return status;
}
public void setStatus(String status) {
    this.status = status;
}
public String getProduct() {
    return product;
}
public void setProduct(String product) {
    this.product = product;
}

}

Then,

   @RequestMapping(value = "/values", method = RequestMethod.POST,produces={"application/json"},
            consumes={"application/json"})
    public ResponseEntity myMethod(
                @RequestBody Order orderObj) {
    }

Please take reference from this example and build the solution.

answered Mar 20, 2017 at 12:14

ProgrammerBoy

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3 Comments

jones j alapat Over a year ago

I cannot change request type from GET

ProgrammerBoy Over a year ago

that not a ideal way to send JSON formatted parameter in URL. Either you have to encode it or use POST method. By any way if you get the GET working then also you will end up in some other issue because what you are expecting is not ideal way to handle JSON.

jones j alapat Over a year ago

yes.I understand your reservations.The thing is that this works seamlessly in our other application in different Technology.So somehow i need to do this in the exactly same way.

Stenal P Jolly · Accepted Answer · 2017-03-20 12:03:03Z

0

You can directly pass String into the JSONObject constructor

There are lots of libraries available.

1.http://www.java2s.com/Code/Jar/j/Downloadjavajsonjar.htm (JSON jar)

2.http://www.java2s.com/Code/Jar/j/Downloadjsonsimple11jar.htm (Simple JSON)

3.http://www.java2s.com/Code/Jar/g/Downloadgson224sourcesjar.htm (GSON from google)

I personally use JSON and GSON jars.

JSONObject jsonObject = new JSONObject(variable)

answered Mar 20, 2017 at 12:03

Stenal P Jolly

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7 Comments

jones j alapat Over a year ago

I am not looking for parsing the JSONString to Object but rather to accept JSON in my query parameter

Stenal P Jolly Over a year ago

Put your RHS of your parameter in quotes and encode before sending to the server. You can decode the text on the server side and convert to any format you wish

jones j alapat Over a year ago

I cannot change that .I could rather change the parameter into JSONObject

slim Over a year ago

What do you mean "I cannot change that". The HTTP standard is the http standard, and if it says you can only use certain chars, that's that: stackoverflow.com/questions/1547899/…

jones j alapat Over a year ago

Http standard does not block the use of above values.Please read the entire q&a

|

Januka samaranyake · Accepted Answer · 2017-03-20 16:41:17Z

0

You can simply try something like this

@GetMapping(value = "/values", produces = MediaType.APPLICATION_JSON_VALUE)
@Timed
public ResponseEntity<String> testMethod(@RequestParam("key") String key) throws Exception {
    JSONObject jsonObject = new JSONObject(key);
    //
}

edited Mar 20, 2017 at 16:41

answered Mar 20, 2017 at 16:20

Januka samaranyake

2,6072 gold badges34 silver badges54 bronze badges

2 Comments

jones j alapat Over a year ago

Throws illegal argument exception for the above URL

Januka samaranyake Over a year ago

@jonesjalapat I think problem with your maven dependencies. Please re check

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