CODEWITHSUNDEEP
javahttpokhttp
John Rock
John Rock

Reputation: 31

Send raw POST request with OkHttp 3

To send a POST request with OkHttp I've seen they use FormBody.Builder, but that requires me to manually input each name and value for the POST data. Is there a way to create a RequestBody with a single String with the POST data? For example, instead of

FormBody.Builder bodyBuilder = new FormBody.Builder();

bodyBuilder.add("name", "value");
bodyBuilder.add("name2", "value2");

...just doing

bodyBuilder.add("name=value&name2=value2");

Upvotes: 3

Views: 4843

Answers (4)

Eray Erdin
Eray Erdin

Reputation: 3179

You can also simply do this with the latest version:

val reqBody = "foo=bar"
val request = Request.Builder().post(reqBody.toRequestBody()).build()

Upvotes: 0

John Dabsky
John Dabsky

Reputation: 33

You can do it like this:

RequestBody body = RequestBody.create(MediaType.get("application/x-www-form-urlencoded"), "key1=value1&key2=value2");

Request request = new Request.Builder()
.url(url)
.post(body)
.build();

Upvotes: 3

rcorreia
rcorreia

Reputation: 559

Create a MediaType variable named FORM:

public static final MediaType FORM = MediaType.parse("multipart/form-data");

Create a RequestBody using the FORM variable and your unparsed params (String):

RequestBody requestBody = RequestBody.create(FORM, params);

Post the full requestBody variable.

Request request = new Request.Builder()
    .url(url)
    .post(requestBody)
    .build();

Upvotes: 0

Antoniossss
Antoniossss

Reputation: 32550

Sure, using RequestBuilder

  Request request = new Request.Builder()
      .url(url)
      .post(body) // here you put your body
      .build();

But URL encoding and other related stuff will be on your side.

Upvotes: 0

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