嘿,伙计们,我的php代码下面是不工作的邮递员,每当我输入电子邮件和密码参数它不工作,并显示只有一个错误下面它显示{成功:0,消息:无效的请求},我给了正确的参数从数据库,我的数据库表名是用户和列是id,名称,电子邮件,密码,类型。请检查代码我是新的在php。请帮帮忙
<?php header('Access-Control-Allow-Origin: *'); ?>
<?php header('Access-Control-Allow-Headers: Origin,X-Requested-With,Content-Type,Accept'); ?>
<?php header('Access-Control-Allow-Methods: POST,GET,OPTIONS,PUT'); ?>
<?php
session_start();
include_once('configdb.php');
error_reporting(E_ALL);
$response = array();
if(isset($_POST['email']) && isset($_POST['password'])){
$name= $_POST['email'];
$password=$_POST['password'];
$result=mysqli_query($conn,"SELECT * FROM users WHERE email='$email' AND password='$password'");
$row=mysqli_fetch_assoc($result);
$count=mysqli_num_rows($result);
if($count==1){
$_SESSION['users']=array(
'email'=>$row['email'],
'password'=>$row['password'],
'name'=>$row['name'],
'type'=>$row['type']
);
$role=$_SESSION['users']['type'];
//Redirecting User Based on Role
switch($role){
case 'user':
// header('location:user.php');
$response["success"] = 1;
$response["message"] = "user.";
break;
case 'admin':
// header('location:admin.php');
$response["success"] = 1;
$response["message"] = "admin.";
break;
}
}else{
$response["success"] = 0;
$response["message"] = "PASSWORD OR EMAIL DOES NOT EXIST.";
}
}
else{
$response["success"] = 0;
$response["message"] = "INVALID REQUEST.";
}
header('Content-type: application/json');
echo json_encode($response);
mysqli_close($conn);
?>
你能告诉我错误是什么吗?
好的,这是我的请求POSTMAN REQUEST
发布于 2018-06-16 04:54:35
您在查询中使用了$email
变量,但您在$name
变量中保存了密码。
https://stackoverflow.com/questions/50882371
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