我正在尝试使用 W3 学校网站的代码通过下拉菜单实现存储在 MySQL 数据库中的数据的简单过滤器(请记住，我对 javaScript 非常陌生!)。但是 Ajax 脚本不返回任何结果。任何帮助表示赞赏。
ajax.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="genre" onchange="showUser(this.value)">
<option value="">Select a genre:</option>
<option value="1">clubbing</option>
<option value="2">comedy</option>
</select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>
</body>
</html>
获取用户.php
<!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>
<script type = "text/javascript" src="ajax.html"></script>
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','','python');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax");
$sql="SELECT * FROM info WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>Venue</th>
<th>Date</th>
<th>Genre</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['venue'] . "</td>";
echo "<td>" . $row['datez'] . "</td>";
echo "<td>" . $row['genre'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
</body>
</html>
最佳答案
找到了解决方案，如果有人遇到类似的问题......
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var gen = document.getElementById('gen').value;
var queryString = "?gen=" + gen ;
ajaxRequest.open("GET", "getuser.php" +
queryString, true);
ajaxRequest.send(null);
}
关于php - 简单的 Ajax 过滤器不返回结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29396773/