我目前正在从事一个几乎完成的项目，但还有一个问题需要解决，这非常关键。
我要做的是在我单击某个图像时更改数据库中的值（activeID）。
我知道我需要 php 和 javascript，但是我的 javascript 技能不是很好，如果有人可以帮助我，那就太棒了！
为了帮助大家理解，这里是数据库中表的结构：
SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;
CREATE TABLE IF NOT EXISTS `artikel` (
`Titel` varchar(100) CHARACTER SET utf8 NOT NULL,
`ArtikelID` int(255) NOT NULL,
`content` text CHARACTER SET utf8 NOT NULL,
`imagepath` varchar(255) CHARACTER SET utf8 NOT NULL,
`activeID` int(11) NOT NULL,
`thumbpath` varchar(255) CHARACTER SET utf8 NOT NULL,
`shortcontent` text CHARACTER SET utf8 NOT NULL,
`categorie` varchar(10) CHARACTER SET utf8 NOT NULL,
PRIMARY KEY (`ArtikelID`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1;
INSERT INTO `artikel` (`Titel`, `ArtikelID`, `content`, `imagepath`, `activeID`, `thumbpath`, `shortcontent`, `categorie`) VALUES
('Zware hinder op E313 door dodelijk ongeval in Geel', 1, 'Het ongeval gebeurde rond 12 uur. Er had zich een file gevormd doordat de afrit Geel-West is afgesloten wegens werken.', 'images/bin1.jpg', 0, 'images/bin1thumb.jpg', 'Op de E313 richting Antwerpen is vandaag ter hoogte van de afrit Geel-West.', 'bin'),
('België geeft 12 miljoen voor heropbouw Afghanistan na 2014', 2, 'Eind 2014 trekt de NAVO zich terug uit Afghanistan. De militairen dragen dan de verantwoordelijkheid voor de veiligheid definitief over aan de Afghaanse veiligheidstroepen.', 'images/bin2.jpg', 0, 'images/bin2thumb.jpg', 'Na de terugtrekking van de Belgische militairen uit Afghanistan eind 2014 zal België voor maximaal 12 miljoen euro bijdragen aan de heropbouw van het land.', 'bin'),
('Oppositie hekelt gebrek aan visie en groei in debat begrotingscontrole', 3, 'De controle was nodig omdat de regering bij de initiële begroting 2012 uitgegaan was van een economische groei van 0,8% van het bbp, terwijl dat cijfer in de economische begroting van februari teruggebracht werd 0,1%.', 'images/bin3.jpg', 0, 'images/bin3thumb.jpg', 'De plenaire Kamer buigt zich vandaag over de begrotingscontrole die het begrotingstekort voor 2012 moet beperken tot 2,8 procent van het bbp, overeenkomstig het stabiliteitsprogramma. Daarvoor was een inspanning van 1,8 miljard euro nodig.', 'bin')
/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;
这是我用于数据库的连接：
<?php
session_start();
$sHost = "localhost";
$sUser = "root";
$sPassword = "";
$sDatabase = "project";
$link = mysqli_connect($sHost, $sUser, $sPassword, $sDatabase) or die("MySQL Error: " . mysqli_error());
?>
这是我必须从数据库中获取内容的 php 和 html 代码，请注意，通过每个记录的路径调用了一个图像（只有您看到的第一个 select 语句与该主题相关，因为我已将其余的 HTML 代码以保持主题简短）：
<?php
//session_start();
//include("classes/connection.php");
include_once("classes/connection.php");
$getBinnenland = mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'bin';");
//$getBuitenland = mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'bui';");
//$getSport = mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'spo';");
//$getPolitiek = mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'pol';");
//$getMilieu= mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'mil';");
//$getBizar = mysqli_query($link,"SELECT * FROM artikel WHERE categorie = 'biz';");
?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Digital Press overview</title>
<link rel="stylesheet" href="style.css" type="text/css" />
</head>
<body>
<div id="wrapper">
<ul id="Binnenland">
<?php
if(mysqli_num_rows($getBinnenland) > 0)
{
while($Binnenland = mysqli_fetch_assoc($getBinnenland))
{
echo "<li><a href='#'><h3>".$Binnenland['Titel']."</h3><img src='". $Binnenland['thumbpath'] ."' alt='thumbnail'/></a><p>".$Binnenland['shortcontent']. "</p>";
}
}
else
{
echo "<li>Nog geen designs</li>";
}
?>
</ul>
</div>
</body>
所以现在的问题：如果我单击某个图像，我希望它将该记录的 activeID 更改为 1 而不是 0，如果我再次单击它，我希望它变回 0
我不知道如何开始这个，所以非常欢迎任何帮助，提前感谢所有帮助我的人！问候
编辑
我已经实现了下面答案中的代码，没有错误，但数据库也没有更改 activeID。有谁知道我能做些什么来解决这个问题？
onclick 事件：
echo "<li><a href='#'><h3>".$Binnenland['Titel']."</h3><img onClick = 'javascript: setActive();' src='". $Binnenland['thumbpath'] ."' alt='thumbnail'/></a><p>".$Binnenland['shortcontent']. "</p>";
的JavaScript：
<script type="text/javascript">
function setActive()
{
if (window.XMLHttpRequest)
{
xmlhttp=new XMLHttpRequest();
}
else
{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
//Return Value. Handle as you wish. Display or ignore.
var x = xmlhttp.responseText;
}
}
xmlhttp.open("POST","setActive.php",true);
xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
xmlhttp.send(p_str);
}
</script>
setActive.php 文件：
<?php
session_start();
$sHost = "localhost";
$sUser = "root";
$sPassword = "";
$sDatabase = "project";
$link = mysqli_connect($sHost, $sUser, $sPassword, $sDatabase) or die("MySQL Error: " . mysqli_error());
$res = mysqli_fetch_row(mysqli_query($link,"select activeID from artikel"));
$active = $res[activeID];
$activeID = ($active == 0) ? 1 : 0;
$res = mysqli_query($link,"update artikel set activeID = $activeID");
if ($res) echo "Success, ActiveID = $activeID";
else echo "Failure, unable to update ActiveID to $activeID";
?>