0
此登录过程在我的本地主机上正常工作,但是当我将其上载到主机上并尝试在实时服务器上测试时,它失败! 我上传了我的数据库到服务器,并试图用注册用户登录,但是当我提交表单时什么都没有发生! 我使用ajax并提醒从服务器返回的数据,它只是提醒一个空白的消息,没有错误,没有警告,没有数据回应!登录表单在本地主机上正常工作,但不在主机上
<?php
$username = $_POST['username'];
$password = $_POST['password'];
$id;
$status = 1;
require_once("../scripts/config.php");
$stmt = $con->stmt_init();
if ($stmt = $con->prepare("SELECT COUNT(email) AS counte FROM users WHERE email=?"))
{
$stmt->bind_param("s", $username);
$stmt->execute();
$obj = $stmt->get_result()->fetch_object();
$counte = $obj->counte;
$stmt->close();
}
if($counte == 1) // email mojud ast
{
if ($stmt = $con->prepare("SELECT COUNT(*) AS countr, id, fname, lname, img FROM users LEFT JOIN userd ON users.id = userd.userid WHERE email=? AND password=?"))
{
$stmt->bind_param("ss", $username, $password);
$stmt->execute();
$obj = $stmt->get_result()->fetch_object();
$countr = $obj->countr;
$id = $obj->id;
$fname = $obj->fname;
$lname = $obj->lname;
$img = $obj->img;
$stmt->close();
if($countr == 0)
{
echo "incorrect";
}
else if($countr == 1)
{
session_start();
$_SESSION['userid'] = $id;
echo $img."*".$fname." ".$lname."*".$id."*";
}
}
}
else if($counte == 0)
{
echo "notexist";
}
mysqli_close($con);
?>
as i sa ID它在我的本地主机上正常工作!我认为问题是与准备好的声明,因为我查了其他查询,我发现他们都没有工作! – behruz
你检查了你的php.ini吗? – KimDev
我不知道如何从c面板访问它! – behruz