CODEWITHSUNDEEP
javaandroidokhttp
freddieptf
freddieptf

Reputation: 2234

Can't get OkHttp's response.body.toString() to return a string

I'm trying to get some json data using OkHttp and can't figure out why when i try logging the response.body().toString() what i get is Results:﹕ com.squareup.okhttp.Call$RealResponseBody@41c16aa8

try {
        URL url = new URL(BaseUrl);
        OkHttpClient client = new OkHttpClient();
        Request request = new Request.Builder()
                .url(url)
                .header(/****/)
                .build();

        Call call = client.newCall(request);
        Response response = call.execute();

        **//for some reason this successfully prints out the response**
        System.out.println("YEAH: " + response.body().string());

        if(!response.isSuccessful()) {
            Log.i("Response code", " " + response.code());
        }

        Log.i("Response code", response.code() + " ");
        String results = response.body().toString();

        Log.i("OkHTTP Results: ", results);

Log

I don't know what i'm doing wrong here. How do i get the response string?

Upvotes: 121

Views: 161462

Answers (10)

Oded Regev
Oded Regev

Reputation: 4405

Just in case someone bumps into the same weird thing as I have. I run my code during development in Debug Mode and apparently since OKHttp 2.4

..the response body is a one-shot value that may be consumed only once

So when in debug there is a call "behind the scene" from the inspector and the body is always empty. See: https://square.github.io/okhttp/2.x/okhttp/com/squareup/okhttp/ResponseBody.html

Upvotes: 55

Nethaji Narasimalu
Nethaji Narasimalu

Reputation: 259

The response.body.string() can be consumed only once. Please use as below:

String responseBodyString = response.body.string();

Use the responseBodyString as needed in your application.

Upvotes: 22

Shubham Vyas
Shubham Vyas

Reputation: 51

Kotlin Programmers I am here for you

 response.body?.charStream()?.readText()?.let { 
                        //Where it refers to Response string
                        callBack.onAPISuccess(it)
                    }

Here you can not use .toString() function and .string() function is not available in Kotlin than you can user charStream() and than convert that charStream into readText() but you have to unwrap the whole value before passing it.But it will never create problem.

I have not explored these charStream() and readText() functions in java but I think it should be there and you can use this in java if these functions are available because I just got to know that java has deprecated the .string() function.

Upvotes: 0

Utkarsh chakrwarti
Utkarsh chakrwarti

Reputation: 11

   Call call = client.newCall(request);
   return call.execute().body().string();

we can get response as a return fromby these

Upvotes: 1

Marko Slijepčević
Marko Slijepčević

Reputation: 443

Instead of using .toString() which returns an object

String results = response.body().toString();

you can use

String results = response.body().string();

Upvotes: 2

Imran Baig
Imran Baig

Reputation: 387

Following is my modified CurlInterceptor. Check the end of the intercept function where I m recreating the Response object after consuming the old Response.

var responseBodyString = responseBody?.string()

response = response.newBuilder() .body( ResponseBody.create( responseBody?.contentType(), responseBodyString.toByteArray() ) ) .build()

class CurlInterceptor: Interceptor
{

    var gson = GsonBuilder().setPrettyPrinting().create()

    override fun intercept(chain: Interceptor.Chain): Response {

    Timber.d(" **** ->>Request to server -> ****")

    val request = chain.request()
    var response = chain.proceed(request)

    var curl = "curl -v -X  ${request.method()}"

    val headers = request.headers()

    for ( i in 0..(headers.size() -1) ){
        curl = "${curl} -H \"${headers.name(i)}: ${headers.value(i)}\""
    }

    val requestBody = request.body()
    if (requestBody != null) {
        val buffer = Buffer()
        requestBody.writeTo(buffer)
        var charset: Charset =
            Charset.forName("UTF-8")
        curl = "${curl} --data '${buffer.readString(charset).replace("\n", "\\n")}'"
    }

    Timber.d("$curl ${request.url()}")
    Timber.d("response status code ${response.code()} message: ${response.message()}")

    
    dumbHeaders(response)

    var responseBody = response?.body()

    if(responseBody != null )
    {
        var responseBodyString = responseBody?.string()

            response = response.newBuilder()
                .body(
                    ResponseBody.create(
                        responseBody?.contentType(),
                        responseBodyString.toByteArray()
                    )
                )
                .build()


        responseBodyString = gson.toJson(responseBodyString)

        Timber.d("response json -> \n $responseBodyString")

    }

    Timber.d(" **** << Response from server ****")

    return response
}



fun dumbHeaders(response: Response) {
    try {
        if (response.headers() != null) {

            for (headerName in response.headers().names()) {
                for (headerValue in response.headers(headerName)) {
                    Timber.d("Header $headerName : $headerValue")
                }
            }
        }
    }
catch (ex: Exception){}
}
}

Upvotes: 3

Imran Baig
Imran Baig

Reputation: 387

Recreate the response object after consuming the string

val responseBodyString = response.body()!!.string()

response = response.newBuilder() .body(ResponseBody.create(responseBody?.contentType(), responseBodyString.toByteArray())) .build()

Upvotes: 0

riddle_me_this
riddle_me_this

Reputation: 9145

Given that a response can potentially produce an OutOfMemoryError in cases of large files, you can instead "peek" the body with the number of bytes and call the string() method.

Note that this will consume the body.

response.peekBody(500).string());

Upvotes: 2

Aoun allah Billel
Aoun allah Billel

Reputation: 47

try to change it like that for example: 

protected String doInBackground(String... params) {
            try {
                JSONObject root = new JSONObject();
                JSONObject data = new JSONObject();
                data.put("type", type);
                data.put("message", message);
                data.put("title", title);
                data.put("image_url", imageUrl);
                data.put("uid",uid);
                data.put("id", id);
                data.put("message_id", messageId);
                data.put("display_name", displayName);
                root.put("data", data);
                root.put("registration_ids", new JSONArray(receipts));
                RequestBody body = RequestBody.create(JSON, root.toString());
                Request request = new Request.Builder()
                        .url(URL)
                        .post(body)
                        .addHeader("Authorization", "key=" + serverKey)
                        .build();
                Response response = mClient.newCall(request).execute();
                String result = response.body().string();
                Log.d(TAG, "Result: " + result);
                return result;
            } catch (Exception ex) {
                Log.e(TAG,"Exception -> "+ex.getMessage());
            }
            return null;
        }

Upvotes: 0

V.J.
V.J.

Reputation: 9580

You have use .string() function to print the response in System.out.println(). But at last in Log.i() you are using .toString().

So please use .string() on response body to print and get your request's response, like:

response.body().string();

NOTE:

  1. .toString(): This returns your object in string format.

  2. .string(): This returns your response.

I think this solve your problem... Right.

Upvotes: 314

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