【发布时间】:2019-11-17 10:33:36
【问题描述】:
我想在 Ajax 中传递数据,但是在运行该函数时,当我检查元素时会出现内部服务器错误。我正在构建一个登录表单,但是当单击登录按钮时,它会出现内部服务器错误。
function login (){
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
console.log("Test2");
document.getElementById("IdForm").innerHTMl = this.responseText;
console.log("Test3");
alert (this.responseText);
}
};
xhttp.open("POST", "PHP/login.php", true);
var data = new FormData(document.getElementById("IdForm"))
console.log("Status500")
xhttp.send(data);
return false;
}
我这里有 PHP 代码:
<?php
$servername = "localhost";
$username = "*****";
$password = "******";
$dbname = "*******";
//create connection
$conn = new mysqli ($servername, $username, $password, $dbname);
//Check connection
if ($conn ->connect_error)
{
die("connection failed:" . $conn -> connect_error);
}
$sql = "SELECT * FROM accounts";
$result = $conn ->query($sql);
$out = "";
$rc = "";
if ($result->num_rows > 0)
{
while($row = $result -> fetch_assoc())
{
echo "<p> " . $_POST ["username"] . "</p>";
if (($row ["username"] == $_POST ["username"]) and $row ["password"] == $_POST ["password"])
header('Location: ok.html');
else
echo "User not valid";
}
} else{
echo "0 results";
}
*/
?>
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