我無法獲取此AJAX代碼來更新我的數據庫。該代碼是一個形象,onClick
將運行命令來更新數據庫使用jQuery使用jQuery更新數據庫
HTML:
<a>
<img class = "heart"
src = "images/heart.png"
onClick = "favUpdate(0,1)"
onMouseover = "this.src='images/heart_mo.png'"
onMouseout = "this.src='images/heart.png'"/>
</a>
Javascript代碼:
function favUpdate(fav_up, id_up) {
$.ajax({
type: 'post',
url: 'includes/fav_update.php',
data: {favorite: fav_up, id: id_up},
success: function(output) {
alert('success, server says '
+ output
+ 'Variables passed are '+fav_up+' '+id_up);
},
error: function() {
alert('something went wrong, Favorite update failed');
}
});
}
PHP代碼:
<?php
require_once('../Connections/main.php');
$fav_update = mysql_real_escape_string($_POST['favorite']);
$fav_id = mysql_real_escape_string($_POST['id']);
$query = "UPDATE projects SET favorite = $fav_update WHERE id = $fav_id";
mysql_query($query, $main);
?>
爲主。 php
<?php
$hostname_main = "localhost";
$database_main = "test";
$username_main = "root";
$password_main = "";
$main = mysql_pconnect($hostname_main, $username_main, $password_main) or trigger_error(mysql_error(),E_USER_ERROR);
?>
有誰知道爲什麼它不更新數據庫,爲什麼「選項」沒有獲取變量的數據?
檢查該查詢運行成功,並告訴我們,如果你得到任何錯誤 –
嘗試'「更新項目SET最喜歡='」。$ fav_update。「'WHERE id ='」。$ fav_id。「'」;' –
您也正在使用obsolesce' mysql_ *'api被棄用,並會導致E_DEPRECATED錯誤php> = 5.5檢查http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php/14110189#14110189 –