[英]XMLHTTP: Internal server error - issue with GET?
我正在尝试从存储在服务器上的MySQL数据库检索一些数据。 我正在使用PHP,Javascript和AJAX来获取数据。
当我在Chrome中运行HTML文件(New.html)并使用开发人员工具查看代码时,它说:
GET http://example.net/Example/getuser.php?q=2500 (内部服务器错误)
showUser @ New.html:31onchange @ New.html:52
我认为这是指xmlhttp.onreadystatechange,并且.send()行旁边有一个红色的X。
<html>
<head>
<title>New</title>
<meta charset="UTF-8">
//Javascript Code
<script>
function showUser(str) {
if (str == " ") {
document.getElementById("txtHINT").innerHTML = " ";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firfox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.reponseText;
}
};
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send(); //LINE 31
}
}
</script>
//CSS for HTML table
<style>
table {
width: 100%;
border-collapse: collapse;
}
table, td, th {
border: 1px solid black;
padding: 5px;
}
th {
text-align: left;
}
</style>
</head>
<!-- Code for Form -->
<body>
<form>
<select name ="users" onchange="showUser(this.value)"> //LINE 51
<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>MySQL Data should go here</b></div>
</body>
</html>
有谁知道如何解决这些问题? 也许需要将.open()放在代码的前面? 或者也许需要某种处理程序?
PHP文件:
<html>
<head>
<title>Latest Attempt</title>
</head>
<?php
$q = intval($_get['q']);
// put your connection code here
$servername = 'localhost';
$username = 'user';
$password = '12345678';
$dbname = 'ajax_demo';
// create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
mysqli_select($conn,"ajax_demo");
$sql="SELECT * FROM my_DB WHERE id = '".$q."'";
$result = mysqli_query($conn,$sql);
echo "<table>
<tr>
<th>FirstName</th>
<th>LastName</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
while ($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>" . $row['Job'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($conn);
?>
</html>
您的代码中有一些拼写错误,如果您修复了这些拼写错误,您的代码将可以正常运行:
在你的JavaScript中
在这条线
document.getElementById("txtHint").innerHTML = this.reponseText;
您的reponseText
拼写错误,应为responseText
因此此行将类似于:
document.getElementById("txtHint").innerHTML = this.responseText;
在你的PHP中
您必须从页面顶部删除所有这些额外的html标签:
<html>
<head>
<title>Latest Attempt</title>
</head>
以及该标签位于页面底部:
</html>
然后您在这一行中:
$q = intval($_get['q']);
您必须输入$ _GET的大写形式,如下所示:
$q = intval($_GET['q']);
最后,mysqli没有mysqli_select()
函数,因此您必须完全删除以下行:
mysqli_select($conn,"ajax_demo");
现在你可以走了:)
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