This question is kind of a duplicate of HTTPRequest.request with sendData, can't seem to get this to work, but I've got some more information now. The goal here is to send a GET request with query parameters attached. I initially tried to send my request as such:
HttpRequest request = new HttpRequest();
request.open("GET", _url, async:true);
request.onError.listen(_onLoadError, onError: _onLoadError);
request.send(sendData);
Where sendData is a String following the normal format for query parameters (?myVariable=2&myOtherVariable=a, etc), since this is the ultimate goal here. The request gets sent, but I never see any additional data (sendData) go with it in any monitoring tools (I'm using Charles). I then tried:
HttpRequest request = new HttpRequest();
request.open("GET", _url + sendData, async:true);
request.onError.listen(_onLoadError, onError: _onLoadError);
request.send();
So now I'm just attaching the query string to the url itself. This works as intended, but is far from elegant. Is there a better solution?
On a GET request you always add the query string to the URL?
When you create the Uri you can pass in a map with the query parameters if you find this more elegant.
Map query = {'xx':'yy', 'zz' : 'ss'};
String url = "http://localhost:8080/myapp/signinService";
Uri uri = new Uri(path: url, queryParameters : query);
HttpRequest.getString(uri.toString().then((HttpRequest req) ...
According to the W3 XMLHttpRequest Specification:
(In the send() method) The argument is ignored if request method is GET or HEAD.
So simple answer to this question is no. sendData cannot be appended to a GET request, this is by XMLHttpRequest specification and not a limitation of Dart.
That said, for requests like this it may be more readable and idiomatic to use HttpRequest.getString
HttpRequest.getString(_url + sendData).then((HttpRequest req) {
// ... Code here
}).catchError(_onLoadError);
If you want to generate a valid URL from url (String) + query parameters (Map) , you can do the following :
Uri uri = Uri.parse(url).replace(queryParameters: parameters);
String finalURL = uri.toString();
url is a String and parameters is a Map
Related
New to the RestAssured and just checking different ways of working with REST APIs. For testing I am using http://dummy.restapiexample.com/api/v1. In this I am trying the GET employee method using RequestSpecification and groovy way but I am getting different response.
My short code is:
RestAssured.baseURI = "http://dummy.restapiexample.com/api/v1";
RequestSpecification request = RestAssured.given();
Response response = request.get("/employee/72100");
System.out.println(response.getBody().asString());
given().baseUri("http://dummy.restapiexample.com/api/v1").get("/employee/72100").then().log().body();
And output I am getting is:
{"id":"72100","employee_name":"mpr51_0280","employee_salary":"123","employee_age":"23","profile_image":""}
<html>
<body>{"id":"72100","employee_name":"mpr51_0280","employee_salary":"123","employee_age":"23","profile_image":""}</body>
</html>
I am not getting why it is returning response with HTML tags. Can anyone explain or give hint to get same response as first call to get method.
NOTE: You may or may not get details for employeeID 72100
You can use any employee ID from response of:
http://dummy.restapiexample.com/api/v1/employees
Because the Body contains it .
You can use below statement if you need only the response
given().baseUri("http://dummy.restapiexample.com/api/v1").get("/employee/72100").then().log();
OR
Response resp = given().baseUri("http://dummy.restapiexample.com/api/v1").get("/employee/72100");
System.out.println(resp.asString());
How to sent POST request with content-type = application/x-www-form-urlencoded.
Given an access code, I am trying to get the AccessToken using POST request where I have all the information set in POST URL so i don't know what to pass as Http content in the postAysnc method.
According to another post For application/x-www-form-urlencoded, the body of the HTTP message sent to the server is essentially one giant query string -- name/value pairs are separated by the ampersand (&), and names are separated from values by the equals symbol (=). An example of this would be:
MyVariableOne=ValueOne&MyVariableTwo=ValueTwo.
So I have similar case where my POST url has all the information as querystring, in that case I don't know what to pass as HttpContent in the postAysnc method since its a mandatory parameter
HttpClient client = new HttpClient();
StringContent queryString = new StringContent(data);
HttpResponseMessage response = await client.PostAsync(new Uri(url), queryString );
Two options:
Continue to use StringContent, but set the content type: new StringContent(data, Encoding.UTF8, "application/x-www-form-urlencoded")
Instead of using StringContent, use FormUrlEncodedContent; note that this takes in a sequence of KeyValuePair<string, string> describing the values (so for example create a dictionary containing { {"MyVariableOne", "ValueOne"}, {"MyVariableTwo", "ValueTwo"} }
I am using rest-assured to test REST APIs, when hitting the url, url has it own encoded parameters.
Response b = given().
relaxedHTTPSValidation().body(gbody).
with().
contentType(ConfigReader.get("application.json")).
then()
.post(url);
It ran successfully but response is empty.
Please tell what can be the reason
This is because, rest-assured will automatically encode the url when you are encoding the url, you need to use method urlEncodingEnabled(false), so that , it will not encode the url again.
Now your code will become
Response res = given()
.urlEncodingEnabled(false)
.relaxedHTTPSValidation()
.body(gbody)
.with()
.contentType(ConfigReader.get("application.json"))
.then()
.post(url);
I'm completely new toSwift. I need to hit a Post Method webservice with NSDictionary parameters & get the JSON response. I tried usingAlamofire & also NSMutableUrlRequest. Nothing seems to workout for me. I either get 'JSON text did not start with array or object and option to allow fragments not set' error or 'Undefined Variable' response from the server. The same service works fine when I try using Objective-C. As I said earlier, I am completely new toSwift & need your assistance.
My base url: http://myofficeit.in/bizfeed/webservices/client.php
Parameter I wanna Pass:
Parameter =
{
UserName = xyz;
deviceModel = iPhone;
deviceToken = "949264bc cd9c6c851ee64cc74db9078770dd7d971618ec20ce91d2e6eb9f155e";
emailid = "xyz#gmail.com";
location = Asia;
userMobileNo = 1234567890;
};
functionName = register;
The code I used for hitting the service is: http://pastebin.com/aaT4uhS7
Thanks
you can use like
let param: [String:AnyObject] = [
"UserName": iPhone,
"deviceToken": "949264bc cd9c6c851ee64cc74db9078770dd7d971618ec20ce91d2e6eb9f155e",
"emailid": "xyz#gmail.com",
"location": Asia,
"userMobileNo": 1234567890
]
Alamofire.request(.POST, "http://myofficeit.in/bizfeed/webservices/client.php/register", parameters: param).responseJSON { (req, res, json, error) in
print(req)
print(res)
print(json)
print(error)
}
for sample request in Alamofire
As broad as your question is, the broad will be my answer:
The first thing to do, is to get a clear idea about the web service API, which also requires a basic knowledge of the HTTP protocol. So, what you need to understand is, what the server expects in HTTP terminology.
You eventually will find out, how the server will expect its "parameters". Note, that there is no term like "parameters" in the HTTP protocol. So, you need to map them into something the HTTP protocol provides.
Most likely, in a POST request, "parameters" are transferred as the body of the HTTP message, as a content-type which is application/x-www-form-urlencoded, multipart/form-data or application/json.
According to the needs of the server, and with your basic knowledge of HTTP and NSURLSession, NSURLComponents etc., you compose the URL and the body of the request, set Content-Type header and possibly other headers and you are ready to go.
How this eventually looks like is given in the answer of #AnbyKarthik, which used Alamofire, and a command that composes a POST request whose parameters are send in the body whose content-type is x-www-form-urlencoded.
I am trying to have a server respond to a request which needs a XML structure. The easiest way I thought would be to create a POST with a string containing the XML, using the Play Framework.
However, I cannot seem to get it to work. I am calling the test with the following code:
Map<String,String> map = new HashMap<String,String>();
map.put("data", xmlString);
Http.Response response = POST("/server/", map);
When on the server checking the parameters it is not in it as it returns false:
params._contains("data")