[英]Sending Object from Server (PHP) to client (Java) using JSON
我想要实现的是从PHP使用JSON发送对象,并将其也作为Java中的对象接收。 每当我使用以下代码时,它都会从Java发出JSON异常,表明String无法转换为JSONObject。 请帮助您。 谢谢
PHP代码
<?php
require_once("/classes/Login.class.php");//Class that connect to the database
$user = $_GET['ballername'];
$pass = $_GET['ballerpassword'];
$log = new Login($user, $pass);
$log->connect();
//header('Content-type: application/json');
$correct = array('success'=> true)
echo json_encode($log);
?>
JAVA代码
public class UserLoginTask extends AsyncTask<Void, Void, Boolean> {
ConnectToServer connectToServer = null;
String urlstring = null;
Campusian campusian;
@Override
protected Boolean doInBackground(Void... params) {
// TODO: attempt authentication against a network service.
urlstring = CampusPalURL.LOGIN_URL +"ballername=" + mUserName +"&ballerpassword="+mPassword;
HttpClient client = new DefaultHttpClient();
HttpGet getMethod = new HttpGet(urlstring);
HttpResponse response;
boolean confirmation;
try {
response = client.execute(getMethod);
HttpEntity httpEntity = response.getEntity();
String state = EntityUtils.toString(httpEntity);
Log.e("STTTTTTTTTTTTTTTTTTTTTTTTTTTT ", state);
try {
JSONObject jsonObject = new JSONObject(state);
//JSONObject confirmation = jsonObject.getJSONObject("");
confirmation = jsonObject.getBoolean("logged_in");
return confirmation;
} catch (JSONException e) {
// TODO Auto-generated catch block
Log.e("Object: ", e.getMessage());
e.printStackTrace();
}
} catch (IOException ee) {
// TODO Auto-generated catch block
Log.e("EXCEPTION: ", ee.getMessage());
ee.printStackTrace();
}
// TODO: register the new account here.
return true;
}
日志错误
01-07 10:54:22.689: E/Object:(1022): Value <!DOCTYPE of type java.lang.String cannot be converted to JSONObject
我决定检查从服务器接收到的输入,该输入在日志中显示EntityUtils.toString(httpEntity),这就是我得到的。 这样对吗?
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <html xmlns="http://www.w3.org/1999/xhtml">
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <head>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <title>Untitled Document</title>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): </head>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <body>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <html xmlns="http://www.w3.org/1999/xhtml">
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <head>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <title>Untitled Document</title>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): </head>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): <body>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): </body>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): </html>{"logged_in":true} </body>
01-07 11:44:37.171: E/STTTTTTTTTTTTTTTTTTTTTTTTTTTT(1380): </html>
您不应该在PHP输出中添加DOCTYPE或HTML标记,而只是回显(或消亡)JSON内容。
小会话示例:
session_start();
$_SESSION['uid'] = $u_id;
$result = mysql_query("UPDATE USER SET USER_SESSION ='".session_id()."' WHERE user_id=$u_id");
$arr = array('Data' => null,'Code' => null);
$arr['Code'] = 200;
$arr['Data']['Session_ID'] = session_id();
echo json_encode($arr);
exit;
我将[Data]用于返回的任何数据,并将[Code]用于错误代码,它非常易于解析和证明。
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