使用POST方法RESTAPI登录android
我是android开发的初学者,正在开发一个android应用程序,使用RESTAPI登录。我必须使用POST方法登录。 在浏览了文档和站点之后,我尝试实现下面的代码,但每次都会发出无效的post请求。我试图调试,但找不到原因。有人可以帮助我的链接,以了解我如何才能实现这一点。 我们必须传递JSON参数{username:abc@test.com,密码:abctest}我想这是一般性的使用POST方法RESTAPI登录android,android,api,rest,post,Android,Api,Rest,Post,我是android开发的初学者,正在开发一个android应用程序,使用RESTAPI登录。我必须使用POST方法登录。 在浏览了文档和站点之后,我尝试实现下面的代码,但每次都会发出无效的post请求。我试图调试,但找不到原因。有人可以帮助我的链接,以了解我如何才能实现这一点。 我们必须传递JSON参数{username:abc@test.com,密码:abctest}我想这是一般性的 如果要将JSON作为唯一参数传递,则必须使用StringEntity而不是UrlEncodedFormEnti
如果要将JSON作为唯一参数传递,则必须使用StringEntity而不是UrlEncodedFormEntity,并发送整个字符串,如下所示:
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://beta.m-adaptive.com/login");
httpPost.setEntity(new StringEntity(YOUR JSON));
httpPost.setHeader("Accept", "application/json");
httpPost.setHeader("Content-type", "application/json");
try {
HttpResponse httpResponse = httpClient.execute(httpPost);
...
这可能有帮助:
class PlaceOrder extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... params) {
// TODO Auto-generated method stub
try {
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPst = new HttpPost(
"yout_url");
ArrayList<NameValuePair> parameters = new ArrayList<NameValuePair>(
2);
parameters.add(new BasicNameValuePair("username", "apple"));
parameters.add(new BasicNameValuePair("pw", "apple"));
parameters.add(new BasicNameValuePair("email",
"apple@gmail.com"));
parameters.add(new BasicNameValuePair("name", "apple"));
httpPst.setEntity(new UrlEncodedFormEntity(parameters));
HttpResponse httpRes = httpClient.execute(httpPst);
String str = convertStreamToString(
httpRes.getEntity().getContent()).toString();
Log.i("mlog", "outfromurl" + str);
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return null;
}
}
public static String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try {
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}
当我点击上面给出的url时,请尝试使用postman客户端查看更多内容。它在broswer中不起作用。请确保它正在运行serviceBasicNameValuePair!=JSONSorry对于延迟回复表示担忧,该链接正在工作,我在AdvanceREST客户端上测试了它,但现在已清除。谢谢你的帮助。
class PlaceOrder extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... params) {
// TODO Auto-generated method stub
try {
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPst = new HttpPost(
"yout_url");
ArrayList<NameValuePair> parameters = new ArrayList<NameValuePair>(
2);
parameters.add(new BasicNameValuePair("username", "apple"));
parameters.add(new BasicNameValuePair("pw", "apple"));
parameters.add(new BasicNameValuePair("email",
"apple@gmail.com"));
parameters.add(new BasicNameValuePair("name", "apple"));
httpPst.setEntity(new UrlEncodedFormEntity(parameters));
HttpResponse httpRes = httpClient.execute(httpPst);
String str = convertStreamToString(
httpRes.getEntity().getContent()).toString();
Log.i("mlog", "outfromurl" + str);
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return null;
}
}
public static String convertStreamToString(InputStream is) {
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try {
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
is.close();
} catch (IOException e) {
e.printStackTrace();
}
}
return sb.toString();
}