[英]Upvote button on website with JavaScript, Php, AJAX, MySql not working. What am I doing wrong?
我正在嘗試從MySql數據庫獲取的文章列表上創建upvote / downvote按鈕。 這些按鈕在您按下按鈕的意義上起作用,並且它會獲得商品的ID。 但是我無法從文章頁面獲取ID到php投票頁面。 當我按下按鈕時,數據庫不會進行投票。 我究竟做錯了什么?
<script type="text/javascript">
$(function() {
$(".vote").click(function()
{
var id = $(this).attr("id");
var name = $(this).attr("name");
var dataString = 'id='+ id ;
var parent = $(this);
if(name=='up')
{
alert('you upvoted on '+ dataString);
$(this).fadeIn(200);
$.ajax({
type: "POST",
url: "weblectureupvote.php",
data: dataString,
cache: false,
});
}
else
{
alert('you downvoted on '+ dataString);
$(this).fadeIn(200);
$.ajax({
type: "POST",
url: "weblectureupvote.php",
data: dataString,
cache: false,
});
}
return false;
});
});
</script>
這是php文件:
<?php
$pid = $_POST['id'];
try {
$db = new PDO('mysql:host=' . $config['db']['host'] . ';dbname=' . $config['db']['dbname'], $config['db']['username'], $config['db']['password']);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$result=mysql_query("SELECT * FROM database WHERE pid = '$pid' ") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
// temp user array
$lecturelist = array();
$lecturelist["pid"] = $row["pid"];
$lecturelist["upvote"] = $row["upvote"];
$lecturelist["downvote"] = $row["downvote"];
$lecturelist["vote"] = $row["vote"];
}
$upvote= $row["upvote"];
$downvote = $row["downvote"];
$vote = $row["vote"];
$upvote = $upvote + 1;
$query = $db->prepare('UPDATE database SET upvote = :upvote WHERE pid = :pid');
$query->execute(array(
':upvote' => $upvote,
':pid' => $pid
));
$query = $db->prepare('UPDATE database SET vote=:vote WHERE pid = :pid');
$query->execute(array(
':vote' => $vote,
':pid' => $pid
));
} catch(PDOException $e) {
echo $e->getMessage();
}
?>
data: {id: id}
這將為您的php文件提供一個“ id”變量(這是第一個id),並帶有一些值(來自第二個id)
現在
$pid = $_POST['id'];
這應該工作,因為您沒有向服務器發送“太多”消息
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