Question

我做了以下教程： http://www.anddev.org/networking-database-problems-f29/connecting-to-mysql-database-t50063.html 使用php将android设备连接到sqlserver（2005）。我检查了我的PHP脚本，它运行并执行正常。当我运行我的程序时，我收到以下错误：

01-26 14:17:43.491: E/log_tag(331): Error parsing data org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONArray
01-27 09:24:13.610: E/log_tag(404): Error parsing data org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONArray
01-27 09:26:45.190: E/log_tag(437): Error parsing data org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONArray
01-27 09:31:14.221: E/log_tag(471): Error parsing data org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONArray
01-27 09:43:44.501: E/log_tag(504): Error parsing data org.json.JSONException: Value Connection of type java.lang.String cannot be converted to JSONArray

我希望我的程序连接到数据库并返回EngineerId大于零的每个人的名字。这是我的代码：

package com.david.DbConnect;


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 DbConnectActivity 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://xxx.xxx.xx.xx/dbconnect.php"; //i use my real ip here



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("EngID","0"));

    //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","ContactName: "+json_data.getInt("ContactName")
                    );
                    //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;
}   

}

这是我的php脚本：

<?php
$serverName = "xxxxxx"; //serverName\instanceName
$connectionInfo = array( "Database"=>"xxxxxx", "UID"=>"xxxxxxxx", "PWD"=>"xxxxxxx");
$conn = sqlsrv_connect( $serverName, $connectionInfo);

if( $conn ) {
     echo "Connection established.<br />";
}else{
     echo "Connection could not be established.<br />";
     die( print_r( sqlsrv_errors(), true));
}

//-----------------------------------------------
// Perform operations with connection.
//-----------------------------------------------
$sql = "SELECT ContactName FROM dbo.TBL_FACILITY_JOB_CALLS WHERE EngineerID>'".$_REQUEST['EngID']."'" ;
$stmt = sqlsrv_query($conn, $sql );
if ($stmt === false) {
     die(print_r(sqlsrv_errors(), true));
}

while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC)) {
      echo $row['ContactName']. "<br />";
}

sqlsrv_free_stmt($stmt);



while($e=sqlsrv_fetch_assoc( $row))

              $output[]=$e;

           print(json_encode($output));

    sqlsrv_close();





?>

有人可以对此有所了解吗？它已经破坏了我的头几天。感谢

Answer 1

PHP输出错误的数据。您不能输出除Json之外的任何数据，因此您应该删除：

 echo "Connection established.<br />";

以及除<：p>之外的任何其他回声数据

print(json_encode($output));

此外，您应该在发送任何数据输出之前为json添加标头：

header('Content-type: application/json');

Answer 2

我认为您从服务器获取的数据不是Json文件的格式。你在这一行有例外：

JSONArray jArray = new JSONArray(result);

确保在结果字符串中，json文件结构中必须有数据。
根据json文件的结构从json中提取数据。

将android设备连接到SqlServer 2005时出现JSON解析错误

2 个答案: