无法使用PHP连接到MySQL从Android使用PHP

Question

我想通过以下教程连接到Android设备的MySQL：http://www.anddev.org/networking-database-problems-f29/connecting-to-mysql-database-t50063.html。

所有这一切发生的是，它正在打印我希望连接到模拟器上的服务器的IP地址。它显示了logcat的以下错误：

12-08 11:42:01.993: I/dalvikvm(274): threadid=3: reacting to signal 3 
12-08 11:42:02.113: I/dalvikvm(274): Wrote stack traces to '/data/anr/traces.txt' 
12-08 11:42:03.273: E/log_tag(274): Error parsing data org.json.JSONException: Value 
<!DOCTYPE of type java.lang.String cannot be converted to JSONArray 
12-08 11:45:58.283: E/log_tag(351): Error parsing data org.json.JSONException: Value 
<!DOCTYPE of type java.lang.String cannot be converted to JSONArray 
    12-08 11:53:11.302: E/log_tag(378): Error parsing data org.json.JSONException: Value 
    <!DOCTYPE of type java.lang.String cannot be converted to JSONObject 
12-08 12:03:31.643: E/log_tag(405): Error parsing data org.json.JSONException: Value 
<!DOCTYPE of type java.lang.String cannot be converted to JSONArray 
12-08 12:20:57.052: E/log_tag(432): Error parsing data org.json.JSONException: Value 
<!DOCTYPE of type java.lang.String cannot be converted to JSONArray 

下面是我的java文件的副本：

package com.david.Connect; 

    import java.io.BufferedReader; 
    import java.io.InputStream; 
    import java.io.InputStreamReader; 
    import java.util.ArrayList; 

    import org.apache.http.HttpEntity; 
    import org.apache.http.HttpResponse; 
    import org.apache.http.NameValuePair; 
    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 org.apache.http.message.BasicNameValuePair; 
    import org.json.JSONArray; 
    import org.json.JSONException; 
    import org.json.JSONObject; 

    import android.app.Activity; 
    import android.os.Bundle; 
    import android.util.Log; 
    import android.widget.LinearLayout; 
    import android.widget.TextView; 


    public class ConnectActivity extends Activity { 
    /** Called when the activity is first created. */ 

     TextView txt; 
    @Override 
    public void onCreate(Bundle savedInstanceState) { 
     super.onCreate(savedInstanceState); 
     setContentView(R.layout.main); 
     // Create a crude view - this should really be set via the layout resources 
     // but since its an example saves declaring them in the XML. 
     LinearLayout rootLayout = new LinearLayout(getApplicationContext()); 
     txt = new TextView(getApplicationContext()); 
     rootLayout.addView(txt); 
     setContentView(rootLayout); 

     // Set the text and call the connect function. 
     txt.setText("Connecting..."); 
     //call the method to run the data retreival 
     txt.setText(getServerData(KEY_121)); 



    } 
    public static final String KEY_121 = "http://86.47.59.249/employee.php"; 



    private String getServerData(String returnString) { 

     InputStream is = null; 

     String result = ""; 
     //the year data to send 
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); 
nameValuePairs.add(new BasicNameValuePair("code","1")); 

//http post 
try{ 
     HttpClient httpclient = new DefaultHttpClient(); 
     HttpPost httppost = new HttpPost(KEY_121); 
     httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 
     HttpResponse response = httpclient.execute(httppost); 
     HttpEntity entity = response.getEntity(); 
     is = entity.getContent(); 

}catch(Exception e){ 
     Log.e("log_tag", "Error in http connection "+e.toString()); 
} 

//convert response to string 
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(); 
     result=sb.toString(); 
}catch(Exception e){ 
     Log.e("log_tag", "Error converting result "+e.toString()); 
} 
//parse json data 
try{ 
     JSONArray jArray = new JSONArray(result); 
     for(int i=0;i<jArray.length();i++){ 
       JSONObject json_data = jArray.getJSONObject(i); 
       Log.i("log_tag","id: "+json_data.getInt("EmployeeId")+ 
         ", name: "+json_data.getString("First_Name")+ 
         ", sex: "+json_data.getInt("Last_Name")+ 
         ", birthyear: "+json_data.getInt("Birth_Date") 
       ); 
       //Get an output to the screen 
       returnString += "\n\t" + jArray.getJSONObject(i); 
     } 
}catch(JSONException e){ 
     Log.e("log_tag", "Error parsing data "+e.toString()); 
} 
return returnString; 
    } 

    } 

，这里是我的PHP文件，我有服务器86.47.59.29在MySQL中数据库是：

<?php 
    mysql_connect("86.47.59.249","username","password"); 
    mysql_select_db("Test"); 

    $q=mysql_query("SELECT * FROM Tbl_Employee WHERE  
    EmployeeId>'".mysql_real_escape_string ($_REQUEST['code'])."'"); 
    while($e=mysql_fetch_assoc($q)) 
    $output[]=$e; 

    print(json_encode($output)); 

    mysql_close(); 
    ?> 

而下面是我的清单文件：

<?xml version="1.0" encoding="utf-8"?> 
<manifest xmlns:android="http://schemas.android.com/apk/res/android" 
    package="com.david.Connect" 
    android:versionCode="1" 
    android:versionName="1.0" > 


    <uses-permission android:name="android.permission.INTERNET"></uses-permission> 

    <application 
     android:icon="@drawable/ic_launcher" 
     android:label="@string/app_name" > 
     <activity 
      android:label="@string/app_name" 
      android:name=".ConnectActivity" > 
      <intent-filter > 
       <action android:name="android.intent.action.MAIN" /> 

       <category android:name="android.intent.category.LAUNCHER" /> 
      </intent-filter> 
     </activity> 
    </application> 
<uses-permission android:name="android.permission.INTERNET"></uses-permission> 
     </manifest> 

任何人都可以对此有所了解吗？它让我疯狂！

感谢

Answer 1

如果我在浏览器中访问http://86.47.59.249/employee.php，我得到一个404错误。这是因为你的PHP脚本要么设置不正确，要么没有正确设置。

你需要做两件事情：

在你的Java，检查请求的响应代码。你可以这样做，像这样：

// ... 
HttpResponse response = httpclient.execute(httppost); 
StatusLine responseStatus = response.getStatusLine(); 
if (responseStatus.getStatusCode() != 200) { 
    // Handle error here 
} else { 
    HttpEntity entity = response.getEntity(); 
    // ... 

在你的PHP，你需要处理潜在的错误：

<?php 

    // Database connection settings 
    $dbHost = '86.47.59.249'; 
    $dbUser = 'username'; 
    $dbPass = 'password'; 
    $dbName = 'Test'; 

    // Try and connect to the database 
    if (!mysql_connect($dbHost, $dbUser, $dbPass)) { 
    header('HTTP/1.1 500 Internal Server Error'); 
    exit('Oh No! Something went wrong connecting to the database: '.mysql_error()); 
    } else if (!mysql_select_db($dbName)) { 
    header('HTTP/1.1 500 Internal Server Error'); 
    exit('Oh No! Something went wrong selecting the database: '.mysql_error()); 
    } 

    // Define SQL query 
    $query = "SELECT * 
      FROM Tbl_Employee 
      WHERE EmployeeId > '".mysql_real_escape_string($_REQUEST['code'])."'"; 

    // Try and execute the query 
    if (!$result = mysql_query($query)) { 
    header('HTTP/1.1 500 Internal Server Error'); 
    exit('Oh No! Something went wrong with the query: '.mysql_error()); 
    } 

    // Fetch all results into an array 
    while ($row = mysql_fetch_assoc($result)) { 
    $output[] = $e; 
    } 

    // Close database link 
    // You can safely leave this line out, PHP implicitly does this when 
    // it terminates 
    mysql_close(); 

    // Exit with a JSON encoded string of the results 
    exit(json_encode($output)); 

Answer 2

在浏览器中打开该页面（employee.php），请查看源代码（CTRL + U大多数浏览器），并确保只返回一个JSON字符串，因为它清楚地在失败试图将结果解析为JSON数组。

Answer 3

使用Firefox的Firebug扩展，看看数据你的PHP脚本返回。你很可能没有使用正确的JSON.get *函数。