I have the following code that I would like to transform into a simple single URL so that (possibly) I can use Picasso for downloading images:
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(Constants.GET_OTHER_PROFILE_PICTURE_URL);
nameValuePairs.add(new BasicNameValuePair(entry.getKey(), entry.getValue()));
UrlEncodedFormEntity form = new UrlEncodedFormEntity(nameValuePairs);
httppost.setEntity(form);
With a GET requests the parameters can be declared in the url like
<url>?<entry.getKey>=<entry.getValue>
The same parameters (as key value pairs separated by a colon) are passed to a POST request through the body.
Related
I need to post my string to server having html content,
Here is my code
Url Builder
Uri.Builder builder = new Uri.Builder()
.scheme("http")
.authority(domain)
.path(pageName)
.appendQueryParameter("myaction", "dopost")
.appendQueryParameter("mesg", msg.replace("\n","<br />"));
And the generated url looks like
domainname/pagename?myaction=dopost&mesg=Hello%2C%3Cbr%20%2F%3E%3Cbr%20%2F%3ETesting
execute http code
HttpGet getRequest = new HttpGet(url);
getRequest.setHeader(CONTENT_TYPE, MIME_TEXT_HTML); // html/text, utf-8
DefaultHttpClient client = new DefaultHttpClient();
HttpResponse response = client.execute(getRequest);
but this will not posted getting forbidden error after executing
Thank in advance
maybe a silly mistake, but a post usually is done with an HttpPost instead of an HttpGet
and looks like this:
// Create a new HttpClient and Post Header
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.yoursite.com/script.php");
try {
// Add your data
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("id", "12345"));
nameValuePairs.add(new BasicNameValuePair("stringdata", "AndDev is Cool!"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
HttpResponse response = httpclient.execute(httppost);
}
I'm using the apache http library and need to know how to add a parameter to an HTTP GET request. I've looked over How to add parameters to a HTTP GET request in Android? but the accepted answer for that adds parameters to an HTTP POST. This is my code so far but it is not working.
HttpGet get = new HttpGet("https://server.com/stuff");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("count", "5"));
HttpParams p = get.getParams();
p.setParameter("length", "5");
get.setParams(p);
String url = "https://server.com/stuff"
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
nameValuePairs.add(new BasicNameValuePair("count", "5"));
HttpClient httpClient = new DefaultHttpClient();
String paramsString = URLEncodedUtils.format(nameValuePairs, "UTF-8");
HttpGet httpGet = new HttpGet(url + "?" + paramsString);
HttpResponse response = httpClient.execute(httpGet);
EDIT: Since Android SDK v22, the type NameValuePair is deprecated. I recommend using Volley, an HTTP library that makes networking for Android apps easier and most importantly, faster.
unlike POST, GET sends the parameters under the url like this:
http://myurl.com?variable1=value&variable2=value2
Where: the parameters area start from the question mark and on so the variable1 is the first param and it has "value" value...
See here for more informations.
So what you need to do is just build an url that contains also these parameters according to server needs.
EDIT:
In your case :
HttpGet get = new HttpGet("https://server.com/stuff?count=5&length=5");
...
Where: count=5 and length=5 are the parameters and the "?" mark is the beginning of the parameters definition...
I hope that helps.
I have a problem in posting the data to server by using httpput methods in android.I have to send feedback to server and getting json response. but i am getting 404 bad request. but i dont know where is the problem.
I am strucked here and didn't find any solution. Any suggestions?
My code is as follows:
HttpClient client = new DefaultHttpClient();
HttpPut put = new HttpPut(getString(R.string.feedBack));
List<NameValuePair> pairs = new ArrayList<NameValuePair>();
pairs.add(new BasicNameValuePair("userId", "8"));
pairs.add(new BasicNameValuePair("feedback",feedbackMessage
.getText().toString()));
put.addHeader("Content-Type", "application/json");
put.addHeader("Accept", "application/json");
put.setEntity(new UrlEncodedFormEntity(pairs));
HttpResponse response = client.execute(put);
Log.d(tag, "Result" + response.getStatusLine());
You are not initializing a URL object and passing in simple String.
You should do this instead:
URL url = new URL(getString(R.string.feedBack));
HttpClient client = new DefaultHttpClient();
HttpPut put= new HttpPut(url);
And it should work.
I just need to send request to webservice via normal HTTP POST inorder to get response.I passed required parameter on body well.While i run it.,i got "Cannot process the message because the content type 'text/json' was not the expected type 'application/soap+msbin1'." error.When i made research over this.,due to "Web Service required the request to have a specific Content-Type, namely "application/soap+msbin1".When i replaced expected content type.,i got Bad Request error.I donno how to recover from that.
My code:
...
DefaultHttpClient httpClient = new DefaultHttpClient();
ResponseHandler <String> resonseHandler = new BasicResponseHandler();
HttpPost postMethod = new HttpPost("My URL");
postMethod.setHeader( "Content-Type", "text/json");
postMethod.setHeader( "Cache-Control", "no-cache");
JSONObject json = new JSONObject();
json.put("userName", "My Username");
json.put("password", "My Password");
json.put("isPersistent",false);
postMethod.setEntity(new ByteArrayEntity(json.toString().getBytes("UTF8")));
HttpResponse response = httpClient.execute(postMethod);
...
It looks like you are trying to call WCF SOAP service. That service expects correct SOAP communication (= no JSON) and moreover it uses MS binary message encoding of SOAP messages (that is what content type describes) is not interoperable so I doubt you will be able to use it on Android device (unless you find implementation of that encoding for Java / Android).
HttpPost request = new HttpPost(url);
StringEntity entity = new StringEntity(json.toString());
entity.setContentType("application/json;charset=UTF-8"); entity.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json;charset=UTF-8"));
request.setHeader("Accept", "application/json");
request.setEntity(entity);
try{
DefaultHttpClient httpClient = new DefaultHttpClient();
response = httpClient.execute(request);
}
Try using something like this. it worked for me.
Thanks.
N_JOY.
I need to access data from a webpage using several different post requests. For now I use:
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("https://myurl");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("action", "search"));
nameValuePairs.add(new BasicNameValuePair("ndc", ndc));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
I need to sent this request using different values for the variable ndc. Would looping this lines be a good idea? If so, how to reuse the HttpClient and HttpPost variables?
If the URL needs to stay the same, then you should only change the values that need to sent.
for (int i=0; i<ndcArray.length;i++)
{
if(i==theNumberWhenURLhasToBeChanged) //adjust this condition based on your knowledge when the url has to be changed, lets say if i > theNumberWhenURLhasToBeChanged, then change the url...
{
httppost = new HttpPost(URLs[theNumberWhenURLhasToBeChanged]);
}
nameValuePairs.add(new BasicNameValuePair("ndc", ndcArray[i]));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
}
Note that: response will change each time, so bear in mind that you should save the response somewhere. And ndcArray[] can be replaced with any structure you want.