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editing this post阐明问题。
4年前关闭。
用最简单的术语来说，我想实现以下目标：
http://www.w3schools.com/php/php_ajax_database.asp唯一的区别是，它不是一个必须从下拉菜单中选择值的选择，而是一个复选框，该复选框将在是否选中该项目时做出反应，如果是，它将调整结果。
我已经奋斗了几个小时，任何帮助将不胜感激。
HTML：
<html>
<head>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form>
<select name="users" onchange="showUser(this.value)">
<option value="">Select a person:</option>
<option value="1">Peter Griffin</option>
<option value="2">Lois Griffin</option>
<option value="3">Joseph Swanson</option>
<option value="4">Glenn Quagmire</option>
</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>
</body>
</html>
PHP文件：
<!DOCTYPE html>
<html>
<head>
<style>
table {
width: 100%;
border-collapse: collapse;
}
table, td, th {
border: 1px solid black;
padding: 5px;
}
th {text-align: left;}
</style>
</head>
<body>
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','peter','abc123','my_db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM user WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "<td>" . $row['Age'] . "</td>";
echo "<td>" . $row['Hometown'] . "</td>";
echo "<td>" . $row['Job'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
更新
<form>
<input type="checkbox" value="1" onchange="showUser(this.value)">
<input type="checkbox" value="2" onchange="showUser(this.value)">
<input type="checkbox" value="3" onchange="showUser(this.value)">
<input type="checkbox" value="4" onchange="showUser(this.value)">
<input type="checkbox" value="4" onchange="showUser(this.value)">
<input type="checkbox" value="5" onchange="showUser(this.value)">
</select>
</form>
您可以使用下面提供的jquery函数serialize在调用ajax时发布所有数据。
data = $( "form" ).serialize();// data can be pass directly to ajax data parameter