I am trying to send request from Java to Twilio SMS API.
I am using java.net.http package:
var url = UriBuilder.fromUri(
"https://api.twilio.com/2010-04-01/Accounts/MyAccount/Messages.json").build();
var urlEncodedBody = URLEncoder.encode(String.format("To=%s&From=%s&Body=%s",
"+1123456789",
"+1223456789",
"Hello"),
StandardCharsets.UTF_8);
var request = HttpRequest.newBuilder(url)
.headers("Authorization", "Basic " + base64,
"Content-Type", "application/x-www-form-urlencoded")
.method("POST", HttpRequest.BodyPublishers.ofString(
urlEncodedBody))
.build();
HttpClient httpClient = HttpClient.newBuilder().build();
try {
var response = httpClient.send(request, HttpResponse.BodyHandlers.ofString());
} catch (IOException |
InterruptedException e) {
throw new RuntimeException(e);
}
I keep getting the error response:
{"code": 21604, "message": "A 'To' phone number is required.", "more_info": "https://www.twilio.com/docs/errors/21604", "status": 400}
Any idea what I'm missing?
I think when you encode the body of the request you are also encoding the &s and =s so it just seems to be one string.
This answer on a different question suggests using a Map and encoding each value in turn. Using that as a basis, your could replace your string formatting with:
Map<String, String> parameters = new HashMap<>();
parameters.put("To", "+1123456789");
parameters.put("From", "+1223456789");
parameters.put("Body", "Hello");
String urlEncodedBody = parameters.entrySet()
.stream()
.map(e -> e.getKey() + "=" + URLEncoder.encode(e.getValue(), StandardCharsets.UTF_8))
.collect(Collectors.joining("&"));
Related
im calling a GET Api (API of ModernMTT) with OkHttp (Last version), my problem is that the reply from the server is not in UTF-8 but in Percent Encoding, and i dont understand why, i try to set MINE in UTF-8 without result.
From CURL and Postman the server reply good.
So the problem is on OkHttp.
My code
public static String translate(String string) throws IOException {
OkHttpClient client = new OkHttpClient().newBuilder().addInterceptor(new FixEncodingInterceptor()).build();
Request request = new Request.Builder()
.url("https://api.modernmt.eu/translate?source=" + sourceLanguage + "&target=" + targetLanguage + "&q=" + string)
.method("GET", null)
.addHeader("MMT-ApiKey", apiKey)
.build();
try (Response response = client.newCall(request).execute()) {
if (!response.isSuccessful()) throw new IOException("Unexpected code " + response);
String stringTranslated= Objects.requireNonNull(response.body()).string();
ResponseMTT responseMTT = new Gson().fromJson(stringTranslated, ResponseMTT.class);
StandardCharsets.UTF_8));
return responseMTT.getData().getTranslation();
}
}
The response.body.string() return this format:
%22Sconto%20del%2020%:%20%3CPAColor0xFF66CC33%3EUtilities%20%E2%86%92%20Capacity%20of%20load%3CPAOldColor%3E.
Can you try to set "Accept-Charset" header to UTF-8 and test your code? I am assuming Postman is defaulting the accept charset header to utf-8.
I'm trying to send a POST request using the new http client api.
Is there a built in way to send parameters formatted as x-www-form-urlencoded ?
My current code:
HttpRequest request = HttpRequest.newBuilder()
.uri(URI.create(url))
.header("Content-Type", "application/x-www-form-urlencoded")
.POST(BodyPublishers.ofString("a=get_account&account=" + URLEncoder.encode(account, "UTF-8")))
.build();
What I'm looking is for a better way to pass the parameters. Something like this:
Params p=new Params();
p.add("a","get_account");
p.add("account",account);
Do I need to build myself this functionality or is something already built in?
I'm using Java 12.
I think the following is the best way to achieve this using Java 11:
Map<String, String> parameters = new HashMap<>();
parameters.put("a", "get_account");
parameters.put("account", account);
String form = parameters.entrySet()
.stream()
.map(e -> e.getKey() + "=" + URLEncoder.encode(e.getValue(), StandardCharsets.UTF_8))
.collect(Collectors.joining("&"));
HttpClient client = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder()
.uri(URI.create(url))
.headers("Content-Type", "application/x-www-form-urlencoded")
.POST(HttpRequest.BodyPublishers.ofString(form))
.build();
HttpResponse<?> response = client.send(request, HttpResponse.BodyHandlers.ofString());
System.out.println(response.statusCode() + " " + response.body().toString());
This way could be useful:
String param = Map.of("param1", "value1", "param2", "value2")
.entrySet()
.stream()
.map(entry -> Stream.of(
URLEncoder.encode(entry.getKey(), UTF_8),
URLEncoder.encode(entry.getValue(), UTF_8))
.collect(Collectors.joining("="))
).collect(Collectors.joining("&"));
You can use up to 10 pairs (param, value) by Map.of(...). It returns an unmodifiable map.
As Łukasz Olszewski said , worked correctly :
String params = Map.of(
Constants.PARAM_CLIENT_ID, apiObject.getClientId(),
Constants.PARAM_SCOPE, apiObject.getScope(),
Constants.PARAM_CODE, apiObject.getCode(),
Constants.PARAM_REDIRECT_URI, apiObject.getRedirectUri(),
Constants.PARAM_GRANT_TYPE, apiObject.getGrantType(),
Constants.PARAM_CODE_VERIFIER, apiObject.getCodeVerifier())
.entrySet()
.stream()
.map(entry -> Stream.of(
URLEncoder.encode(entry.getKey(), StandardCharsets.UTF_8),
URLEncoder.encode(entry.getValue(), StandardCharsets.UTF_8))
.collect(Collectors.joining("="))
).collect(Collectors.joining("&"));
HttpResponse<?> response = utils.consumeHttpPostFormUrlEncodedClientByRequestUrl(Constants.URL_BASE + Constants.URL_GET_TOKEN, params);
and consumeHttpPostFormUrlEncodedClientByRequestUrl
public HttpResponse<?> consumeHttpPostFormUrlEncodedClientByRequestUrl(String url, String map) throws IOException, InterruptedException {
HttpClient httpClient = HttpClient.newHttpClient();
HttpRequest request = HttpRequest.newBuilder(URI.create(url))
.header("Content-Type", String.valueOf(MediaType.APPLICATION_FORM_URLENCODED))
.POST(HttpRequest.BodyPublishers.ofString(map))
.build();
return httpClient.send(request, HttpResponse.BodyHandlers.ofString());
}
Instead of Stream.of you can use more compact String.join (according to Łukasz Olszewski answer):
String form = Map.of("param1", "value1", "param2", "value2")
.entrySet()
.stream()
.map(entry -> String.join("=",
URLEncoder.encode(entry.getKey().toString(), StandardCharsets.UTF_8),
URLEncoder.encode(entry.getValue().toString(), StandardCharsets.UTF_8)))
.collect(Collectors.joining("&"));
return HttpRequest.BodyPublishers.ofString(form);
Check out Methanol. It's got a nice FormBodyPublisher for x-www-form-urlencoded bodies.
var formBody = FormBodyPublisher.newBuilder()
.query("a", "get_account")
.query("account", account)
.build();
var request = MutableRequest.POST("https://example.com", formBody);
// Methanol implements an HttpClient that does nice things to your request/response.
// Here, the Content-Type header will be added for you.
var client = Methanol.create();
var response = client.send(request, BodyHandlers.ofString());
I have a rest post call. whenever I try to hit using postman it is working fine. but If I try same post call using JerseyAPI Client. I'm getting 400 Bad request
String URI = "rest uri";
Client client = Client.create();
WebResource webResource = client.resource(URI);
try {
String input1 = "{\"callType\": \"UPDATE\",\"emails\": [\"qa_tester2222#gmail.com\",\"qa_tester2222#gmail.com\"],\"event\": \"UPDATE_EVENT\",\"externalIds\": [ \"id\" ], \"fraudAction\": \"CONFIRMED_FRAUD\",\"fraudCategory\": \"Account Takeover\",\"memo\": \"test fraud case management for AD\",\"origin\": \"SE4\",\"phones\": [],\"requester\": \"corsairUser\",\"source\": \"LIVE\" }";
ClientResponse response = webResource.type("application/json")
.header("Authorization", "Basic secretKey")
.post(ClientResponse.class, input1);
if (response.getStatus() != 201) {
throw new RuntimeException("Failed : HTTP error code : " + response.getStatus());
}
System.out.println("Output from Server .... \n");
String output = response.getEntity(String.class);
System.out.println(output);
} catch (
Exception e) {
e.printStackTrace();
}
}
If I run same request in Postman using a above input, I'm able to hit api successfully and get 200, but getting 400 in JerseyAPI
Since you are getting a Bad Request response, I think you should create your JSON string using a proper JSON tool. You can start by creating a JSON object from your string and then use the JSON object to retrieve the input.
Creating the JSON object:
JSONObject jsonObject = new JSONObject(string);
Using the JSON object as your input:
ClientResponse response = webResource.type("application/json")
.header("Authorization", "Basic secretKey")
.post(ClientResponse.class, jsonObject.toString());
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);
I don't know what i'm doing wrong here. How do i get the response string?
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:
.toString(): This returns your object in string format.
.string(): This returns your response.
I think this solve your problem... Right.
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/3.x/okhttp/okhttp3/ResponseBody.html
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.
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());
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){}
}
}
Instead of using .toString() which returns an object
String results = response.body().toString();
you can use
String results = response.body().string();
Call call = client.newCall(request);
return call.execute().body().string();
we can get response as a return fromby these
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;
}
Recreate the response object after consuming the string
val responseBodyString = response.body()!!.string()
response = response.newBuilder()
.body(ResponseBody.create(responseBody?.contentType(), responseBodyString.toByteArray()))
.build()
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.
I am struggling with creating a plain text file on a server via HTTP PUT. I am using apache commons httpClient. My credentials are working but there is no body content in my request. What must I do to create the file like this? It works as intended when I try via hurl.it (ie setting my credentials, and setting a body). What I would like is the string "hej" to show in the file body. After getting this to work I intend to use a JSONString. The following code generates an empty file on the server (204 response):
HttpClient httpClient = new DefaultHttpClient();
String encoding = http_username + ":" + http_password;
encoding = Base64.encodeBase64String(encoding.getBytes());
HttpPut httpput = new HttpPut(http_path);
HttpEntity content=null;
try{
content = new StringEntity("hej");
}
catch(UnsupportedEncodingException e){
logger.error("Failed to Encode result");
}
logger.info("executing request " + httpput.getRequestLine());
try {
httpput.setHeader("Authorization", "Basic " + encoding);
//httpput.setHeader("Content-Type", "application/json; charset=utf-8");
httpput.setEntity(content);
HttpResponse response = httpClient.execute(httpput);
Header[] allHeaders = response.getAllHeaders();
for (Header h : allHeaders) {
logger.info(h.getName() + ": " + h.getValue());
}
} catch (Exception e) {
logger.error(e.getMessage());
}
I have tried both setting a content type and not doing it, no difference. What basic thing am I doing wrong?
Turns out that Base64.encodeBase64String appends a newline character at the end of the string, which throws everything off!
String encoding = http_username + ":" + http_password;
encoding = Base64.encodeBase64String(encoding.getBytes());
encoding= encoding.replace("\r\n", ""); //This fixes everything
Wow, that just took me a couple of days to figure out!