Question

我无法获取此AJAX代码来更新我的数据库。代码是onClick将运行命令以更新数据库的图像

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代码：

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代码：

<?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); 
?>

main.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); 
?>

有谁知道为什么它没有更新数据库以及为什么＆＃34;选项＆＃34;是不是要获取变量的数据？

Answer 1

试试这个

<?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); 
?>

Answer 2

如果数据类型为字符串$fav_update，则必须在VARCHAR,TEXT附近放置单引号

$query = "UPDATE projects SET favorite = '$fav_update' WHERE id = $fav_id";
                                         ^           ^

从此处删除$main并尝试

mysql_query($query);

Answer 3

请尝试调试PHP（服务器端代码优先）： -

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_REQUEST['favorite']);
    $fav_id = mysql_real_escape_string($_REQUEST['id']);
    $query = "UPDATE projects SET favorite = $fav_update WHERE id = $fav_id";
    echo $query;
     mysql_query($query, $main); 
?>

使用$ _REQUEST而不是$ _POST，并通过创建类似http://localhost/filename.php?favorite=somevalue1&id=somevalue2的网址

直接从浏览器调用此API

检查是否在DB中插入了插件，并通过打印来检查查询。

检查API后，请将$ _REQUEST更改回$ _POST

Answer 4

将php脚本更改为：

<?php
    require_once('../Connections/main.php');
    $fav_update = mysql_real_escape_string($_POST['favorite']);
    $fav_id = mysql_real_escape_string($_POST['id']);
    $updateSQL = sprintf("UPDATE projects 
                          SET favorite=%s  
                          WHERE id=%s",
                          $fav_update,
                          $fav_id);
    mysql_select_db($database_main, $main);
    $Result1 = mysql_query($updateSQL, $main) or die(mysql_error());


?>

希望如果有人遇到同样的问题会有帮助

使用AJAX使用jQuery更新数据库

4 个答案: