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解析JSON时出错

[英]Error while parsing JSON

I am new to Android development. 我是Android开发的新手。 I'm sending JSON data from Android to my PHP server. 我正在将JSON数据从Android发送到我的PHP服务器。 But I'm getting an error: 但是我遇到一个错误:

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

This is my PHP code: 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)));

?>

Sounds like you've got some HTML embedded, probably a PHP error/warning string. 听起来您已经嵌入了一些HTML,可能是PHP错误/警告字符串。 The response from the server must consist of ONLY the json data. 来自服务器的响应必须包含json数据。 Anything else will become part of the string and cause parse errors. 其他任何内容都将成为字符串的一部分并引起解析错误。

Take the exact url you're hitting in android and see what it brings up in a browser. 取得您在android中命中的确切网址,然后查看它在浏览器中显示的内容。

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