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在我的应用程序中，我想通过 httpPost 中的 url 传递一个 json 参数。我的网址是

http:\\xyz\login.php?

我有 json 参数{"userName"="ekant","password"="xyz"}

我希望这个 json 像http:\\xyz\login.php?userName=ekant&password=xyz查询字符串一样与 url 一起传递。谁能告诉我该怎么做？谢谢

4

5 回答 5

0

你可以使用这个类，我在很多应用程序中都使用过它。

package com.your.package;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.List;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.StatusLine;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;

import android.util.Log;

public class RestPost {
    String url;
    List<NameValuePair> nameValuePairs;

    public RestPost(String str, List<NameValuePair> params) {
        this.url = str;
        this.nameValuePairs = params;
    }

    public String postData() {
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(this.url);
        StringBuilder builder = new StringBuilder();

        try {
            httppost.setEntity(new UrlEncodedFormEntity(this.nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            StatusLine statusLine = response.getStatusLine();
            int statusCode = statusLine.getStatusCode();
            Log.d("RestClient", "Status Code : " + statusCode);

            HttpEntity entity = response.getEntity();
            InputStream content = entity.getContent();
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    content));
            String line;
            while ((line = reader.readLine()) != null) {
                builder.append(line);
            }

        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
        } catch (IOException e) {
            // TODO Auto-generated catch block
        }
        return builder.toString();
    }
}

要使用这个...

List<NameValuePair> pairs = new ArrayList<NameValuePair>();
pairs.add(new BasicNameValuePair("user_id", uid));
pairs.add(new BasicNameValuePair("auth_token", auth));
pairs.add(new BasicNameValuePair("topic_name", TName));

RestPost post = new RestPost(Constants.ForumAddTopic, pairs);
String Response = post.postData();

希望能帮助到你...

于 2013-10-17T04:48:15.177 回答
0

您可以使用 droidQuery 轻松发送：

$.ajax(new AjaxOptions().url("http://www.example.com")//switch with real URL
                        .type("POST")
                        .dataType("json")
                        .data(jsonParams)//your JSON Object
                        .context(this)
                        .success(new Function() {
                            @Override
                            public void invoke($ droidQuery, Object... params) {
                                //if you are expecting a JSONObject:
                                JSONObject response = (JSONObject) params[0];
                                //TODO: handle response
                            }
                        })
                        .error(new Function() {
                            @Override
                            public void invoke($ droidQuery, Object... params) {
                                AjaxError error = (AjaxError) params[0];
                                $.toast(error.status + ": " + error.reason, Toast.LENGTH_LONG);
                                //retry once
                                $.ajax(error.request, error.options.error($.noop()));
                            }
                        }));
于 2013-10-17T04:47:22.177 回答
0

使用名称值对

private static String loginURL = "http://www.xyz.com/login.php";

public JSONObject loginUser(String email, String password) {
    // Building Parameters
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("username", name));
    params.add(new BasicNameValuePair("password", password));
    JSONObject json = jsonParser.getJSONFromUrl(loginURL, params);
    return json;
}
于 2013-10-17T04:43:16.137 回答
0

1.首先将java对象转换为json

JSONObject json = new JSONObject();
   json.put("userName", youvalue);
   json.put("password", yourvlaue);

2.然后将json对象作为参数传递给你的post方法

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", json.toString());
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

    // Execute HTTP Post Request
    HttpResponse response = httpclient.execute(httppost);

} catch (ClientProtocolException e) {
    // TODO Auto-generated catch block
} catch (IOException e) {
    // TODO Auto-generated catch block
}

注意：-此代码未经测试...

于 2013-10-17T04:43:28.387 回答
0

如果你使用 jquery，很容易 slove data = {"userName"="ekant","password"="xyz"}; $.param（数据）。

于 2013-10-17T04:45:45.867 回答