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

转载 作者:行者123 更新时间:2023-11-29 18:14:10 25 4
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我完成了以下教程: http://www.anddev.org/networking-database-problems-f29/connecting-to-mysql-database-t50063.html用于使用 php 将 android 设备连接到 sqlserver(2005)。我检查了我的 php 脚本,它运行良好。当我运行我的程序时,出现以下错误:

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 大于零的每个人的姓名。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();





?>

任何人都可以阐明这一点吗?这几天一直让我头疼。谢谢

最佳答案

PHP 输出了错误的数据。除了Json你不能输出任何数据,所以你应该删除:

 echo "Connection established.<br />";

和任何其他回显的数据,除了:

print(json_encode($output));

此外,在发送任何数据输出之前,您应该为 json 添加 header :

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

关于php - 将 android 设备连接到 SqlServer 2005 时出现 JSON 解析错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9031499/

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