解析JSON时出错

Question

我是Android开发新手。我将Android的JSON数据发送到我的PHP服务器。但我发现了一个错误：

Error parsing data org.json.JSONException: Value `<br` of type java.lang.String cannot be converted to JSONObject". 

这是我的PHP代码：

<?php 

$con = mysql_connect("localhost","custome234r","reswtdf123"); 
if (!$con) 
    die('Could not connect: ' . mysql_error()); 
mysql_select_db("customer_dd_test", $con); 

$jsonFeedbackResult = $_REQUEST['results']; 

$flagToOpenTicket = false;    

$arrResult = json_decode(stripslashes_deep($jsonFeedbackResult)); 

$feedbackname = $arrResult[0]['feedbackname']; 
$email = $arrResult[0]['email']; 

unset($arrResult[0]); 
$finalArray = array_values($arrResult); 
foreach($finalArray as $key => $arrQuestionWithAnswer) 
{ 
    if($arrQuestionWithAnswer['answer'] == 'bad' || $arrQuestionWithAnswer['answer'] == 'worst') 
    { 
     $flagToOpenTicket = true; 
     break; 
    } 
} 

if($flagToOpenTicket) 
{ 
    $insertQuery = 'INSERT INTO dev_ticket(email, feedbackname) VALUES'; 
    $insertQuery .= '("'.$email.'", "'.$feedbackname.'"),'; 
    $executeQuery = trim($insertQuery,','); 
    mysql_query($executeQuery); 

} 
mysql_close($con); 
print(json_encode(array('response'=>$feedbackname))); 

?> 

Answer 1

听起来像是你有一些HTML嵌入，可能是一个PHP错误/警告字符串。来自服务器的响应必须包含仅限的json数据。其他任何东西都会成为字符串的一部分并导致解析错误。

以您在android中击中的确切网址并查看它在浏览器中显示的内容为准。