我想使用ajax成功响应验证登录表单是否有效,但如果所有商品都应该重定向到索引页面,但它不起作用,我不明白出了什么问题,请帮忙。提前致谢..
$(document).ready(function(){
var response;
$('#submit').click(function(e){
e.preventDefault();
$('.alert-box').html('<div id="loading" style="margin: 0 auto;" >
</div>');
var action = 'ajax_validation';
var username = $('#username').val();
var password = $('#password').val();
$.ajax({
url:"do_login.php",
method:"POST",
data:{action:action, username:username, password:password},
success:function(data){
response = data;
if(response === 1){
window.location.href = "http://stackoverflow.com";
}else{
$('.alert-box').addClass("alert alert-warning");
$('.alert-box').html(response);
}
}
});
});
});
上面这些ajax请求 这是操作页面代码 include('includes/db.php');
if(isset($_POST["action"])){
$check_username = $_POST['username'];
$check_password = $_POST['password'];
if(empty($check_username) && empty($check_password)){
echo "Please fill all field";
}else{
$query = "SELECT * FROM user WHERE email = '{$check_username}' AND password = '{$check_password}' ";
$select_query=mysqli_query($connection,$query);
if(!$select_query){
die("QUERY FAILED".mysqli_error($select_query));
}
if(mysqli_num_rows($select_query)==0){
echo "Username or password are incorrect!";
}else{
while ($row=mysqli_fetch_assoc($select_query)) {
$_SESSION['username'] = $row['email'];
echo $row['email'];
}
}
}
}
最佳答案
在你的回复中你正在呼应
echo $row['email'];
这应该是:
echo 1;
关于php - ajax登录成功数据但不重定向到索引页面,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55045794/