Question

现在我有了这个：

$data = array();
$query = mysql_query("SELECT * FROM users WHERE username = '{$username}'");

while ($row = mysql_fetch_row($query)) {
$data[] = $row;
}

现在$data[0] = array('Joe','Sally','3 5'); $data[1] = array('Joe','alice','30 65');

访问$data[0][0] = 'Joe'或$data[1][0] = 'Joe'. $data[1][1] = 'alice'。

然后我如何从php发送这个2D数组并使用JSONObject或JSONArray在android中接收它？

Answer 1

试试这个

$data = array();
$query = mysql_query("SELECT * FROM users WHERE username = '{$username}'");

while ($row = mysql_fetch_row($query)) {
$data[] = $row;
}
 $json=json_encode(array("userlist" => $data));
        echo $json;

Answer 2

用户json_encode($dd)显示Json字符串

$data = array();
$query = mysql_query("SELECT * FROM users WHERE username = '{$username}'");

while ($row = mysql_fetch_row($query)) {
$data[] = $row;
}

echo json_encode（$ data）;

Answer 3

你可以使用json_encode()函数将你的数组转换为json数组。在你的android开发代码中，你必须添加JSONParser来提取你的信息。试试这个java类来从PHP获取信息JSON。

In PHP
$data = array();
$query = mysql_query("SELECT * FROM users WHERE username = '{$username}'");

while ($row = mysql_fetch_row($query)) {
$data[] = $row;
}

echo json_encode($data);

JSONParser.java

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

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {

    }

    // function get json from url
    // by making HTTP POST or GET mehtod
    public JSONObject makeHttpRequest(String url, String method,
            List<NameValuePair> params) {

        // Making HTTP request
        try {

            // check for request method
            if(method == "POST"){
                // request method is POST
                // defaultHttpClient
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);
                httpPost.setEntity(new UrlEncodedFormEntity(params));

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();

            }else if(method == "GET"){
                // request method is GET
                DefaultHttpClient httpClient = new DefaultHttpClient();
                String paramString = URLEncodedUtils.format(params, "utf-8");
                url += "?" + paramString;
                HttpGet httpGet = new HttpGet(url);

                HttpResponse httpResponse = httpClient.execute(httpGet);
                HttpEntity httpEntity = httpResponse.getEntity();
                is = httpEntity.getContent();
            }           

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } 
        catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}

从php发送数组并在android中接收它

3 个答案: