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我有以下设置为我的ajax调用,但它不工作时,我点击一个明星(这是一个星级评分),虽然如果我访问ajax.php直接插入,但ajax调用没有发生。试图使ajax调用工作与PHP和MySQL
在PHP文件的HTML
echo '<table>
<tr>
<td style="padding:10px;">
<input type="hidden" name="userID" value="'.$user_id.'">
<span style="font-size: 20px; vertical-align:top;">Comments</span>
</td>
<td style="padding:10px;">
<textarea name="comments" cols="60" rows="2"></textarea>
</td>
<td>
<div>
<input name="star1000" value "1" type="radio" class="star"/>
<input name="star1000" value="2" type="radio" class="star"/>
<input name="star1000" value="3" type="radio" class="star"/>
<input name="star1000" value="4" type="radio" class="star"/>
<input name="star1000" value="5" type="radio" class="star"/>
</div>
</td>
<tr>
</table>';
JS - 用于发送的值
<script>
$('.star').rating({
callback: function(value, link) {
var name = $(this).attr('name');
var userID = $(this).closest('td').find('input[name="userID"]').val();
var comments = $(this).closest('td').find('textarea[name="comments"]').val();
$.ajax({
url: "http://localhost/mywebsite/ajax.php",
type: "POST",
data: {
name: name,
value: value,
userID: userID,
comments: comments
},
cache: false,
success: function(response) {
try {
console.log(response);
} catch (err) {
alert(response);
}
}
});
}
});
</script>
位于http://localhost/mywebsite/ajax.php
<?php
extract($_POST);
$rate_val = $_POST['value'];
$user_id = $_POST['userID'];
$comments = $_POST['comments'];
$insert_q = "INSERT INTO ratings (rate_comments, rate_num, option_id, rate_date,user_id)
values ('$comments','$rate_val','1',now(),'$user_id')";
include 'opendbconn.php';
if(!($result = mysql_query($insert_q, $database)))
{
print("Could not execute query!<br/>");
die(mysql_error()."</div>
</div>
</body>
</html>");
}
include 'closedbcon.php';
?>
什么事件您呼叫的评价功能 –
当我在“明星”点击它没有被调用?这里它带着你的类 – user710502
你检查过萤火虫吗?它显示任何错误? –