我有一个 php 脚本，我希望用 ajax 调用它，我遇到了来自 ajax 的错误，显示“Error:email=my_email&password=myPassword”。
这是 PHP 脚本
<?php
session_start(); //starts a session in order to be-able to save session variables and to read them
require "db_config.php"; //Allows us to use the database connection from db_config.php in this file
if ($_SERVER["request_method"] == "post"){ //checks if the form was submitted
$email = $_POST["email"]; //fetching the email address which was inserted in the login.html form
$password = $_POST["password"]; //fetching the password which was inserted in the login.html form
/*
querying the database, to check whether there is a result with the email and password entered by the user
*/
$checkForUser = mysqli_query($db_connection, "SELECT * FROM `tbl_users` WHERE email = '$email' and Password = '$password' LIMIT 1");
/*
checking if the query resulted in one row, if there is a row
it means there is a user with this email and password, which means these are the correct creadentials
*/
$rows = mysqli_num_rows($db_connection, $checkForUser);
if ($rows == 1){
//this means: correct credentials
//the next few lines fetch the information from the result
while($row = mysqli_fetch_assoc($checkForUser)){
$_SESSION["user_id"] = $row["userId"]; //creates a session variable containing the users id
$_SESSION["users_name"] = $row["firstName"]. " ".$row["lastName"]; //creates a session variable containing the users name
}
echo "You are now logged in: ". $_SESSION["users_name"];
}else{
//this means: incorrect credentials
echo "Incorrect Username or password"; //prints out error message
}
}
?>
这是main.js
$(document).ready(function() {
$('#loginForm').submit(function() {
var data = $(this).serialize();
$.ajax({
url: "../php/login.php",
type: "POST",
data: data,
success: function(data) {
$('*').html(data);
},
error: function() {
alert('ERROR: ' + data);
}
});
return false;
});
});
这是login.html页面，可能会有帮助
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/main.css">
<link rel="stylesheet" type="text/css" href="css/responsive.css">
<script src="js/jquery.js"></script>
<script src="js/main.js"></script>
<meta name="copyright" content="Yudi Moszkowski">
<title>Login | Your Site Name</title>
</head>
<body>
<div id="loginContainer">
<div class="logo"><img src="img/yourLogo.png"></div>
<form action="php/login.php" method="post" id="loginForm">
<input type="text" name="email" placeholder="Email" id="loginEmail" class="loginInput" required="true">
<input type="password" name="password" placeholder="Password" id="loginPassword" class="loginInput" required="true"/>
<input type="submit" value="Login" name="loginSubmit" id="loginSubmit">
</form>
<div id="loginOptions"><p id="noAccount">Not signed up? <a href="signup.html">Signup</a></p><p id="forgotPass"><a href="forgot_pass.html">Forgot password?</a></p></div>
</div>
</body>
</html>
感谢您的宝贵时间:)
最佳答案
您似乎输入了错误的网址: 在 html 操作中，您使用“php/login.php”，在 ajax 调用中，您使用相同的 url，前面带有“../”。如果您解释一下login.php和这个html文件的位置，这将有助于解决您的问题。
关于javascript - 从 Ajax 调用返回 php 错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37260761/