[英]PHP/AJAX parse error
我正在嘗試使用我的PHP shoutbox和使用AJAX查詢,以便頁面不必重新加載以發布呼喊。 任何幫助將不勝感激!
我被困 - 我一直收到一個解析錯誤:
18:47:27.216 Details: parsererror
Error:SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data 1 index.php:100:4
我有shoutBox.php,它包含HTML和AJAX:
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</head>
<script type="text/javascript">
function sendChatMessage(msg)
{
$.ajax({
type: "POST",
url: "login/postData.php",
data: {
thisPost: msg
},
dataType: "json",
success: function(data)
{
if (data.error === false)
{
$("#aData").text(data);
console.log("Post submitted.");
ajax.reload();
}
else
{
console.log("Post was not submitted.");
}
},
error: function(xhr, desc, err)
{
console.log(xhr);
console.log("Details: " + desc + "\nError:" + err);
}
});
}
</script>
<div class="shoutbox">
<h1 class="panel-heading"><strong>Federal</strong> Communications Array</h1>
<hr>
<ul class="shoutbox-content"></ul>
<div class="panel">
<form method="post">
<p>
<table border="0" align="center" style="height: auto;" width="90%">
<tbody>
<tr>
<td><a></a></td>
<td>
<div class="input-group">
<span class="input-group-addon">Say:</span>
<input class="form-control" id="aData" name="post" maxlength='255' placeholder="Enter a message..."></input>
<div class="input-group-btn">
<button class="btn btn-success" onClick="sendChatMessage(document.getElementById('aData').value); return false;">Send</button>
</div>
</div>
</td>
<td></td>
</tr>
</tbody>
</table>
<p>
</form>
</div>
<div class="panel">
<table class="table table-striped">
<thead>
<tr>
<th width="100px">Timestamp</th>
<th width="120px">Username</th>
<th width="980px">Message</th>
<th></th>
</tr>
</thead>
</table>
</div>
</div>
和postData.php,它(應該)執行查詢。
<?php
ini_set('display_errors', 'On');
error_reporting(E_ALL | E_STRICT);
echo('Script begin.');
require "dbconf.php"; // db details
$connect = mysql_connect($host,$username,$password);
mysql_select_db($db_name,$connect);
$post = '';
if(empty($_POST['aData']))
{
exit(json_encode([
'error' => 'You must enter a message!',
]));
} else {
$post = trim($_POST['aData']);
$playerUsername = $_SESSION['username'];
$idQuery = "SELECT `id` FROM `members` WHERE `username`='$playerUsername'";
$playeridQuery = mysql_query($idQuery);
$playerID = mysql_result($playeridQuery, 0);
mysql_query("INSERT INTO `shoutboxPosts` SET `id`='$playerID', `username`='$playerUsername', `post`='$post'");
}
exit(json_encode([
'error' => false,
]));
?>
刪除一行
<?php
ini_set('display_errors', 'On');
error_reporting(E_ALL | E_STRICT);
//echo('Script begin.'); Remove this line. It's violating JSON format
require "dbconf.php"; // db details
$connect = mysql_connect($host,$username,$password);
mysql_select_db($db_name,$connect);
$post = '';
if(empty($_POST['aData']))
{
exit(json_encode([
'error' => 'You must enter a message!',
]));
} else {
$post = trim($_POST['aData']);
$playerUsername = $_SESSION['username'];
$idQuery = "SELECT `id` FROM `members` WHERE `username`='$playerUsername'";
$playeridQuery = mysql_query($idQuery);
$playerID = mysql_result($playeridQuery, 0);
mysql_query("INSERT INTO `shoutboxPosts` SET `id`='$playerID', `username`='$playerUsername', `post`='$post'");
}
exit(json_encode([
'error' => false,
]));
?>
您的解析錯誤來自此行:
$( “#ADATA”)文本(數據)。
您不能將Json編碼數據發送到DOM中的id。 你想看到數據的價值,試試這個:
警報(JSON.stringify(數據));
其他的東西,在你的PHP,避免退出,忘了if和else。
正確的方法是嘗試,抓住。 你嘗試,如果發生錯誤,捕獲異常。
在函數外部放置一個全局變量並將awswer放入其中。
之后:
echo json_encode($ aswer); $ connect = null; //總是關閉你的連接。
我希望它會對你有所幫助。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.