过程php和sql登录脚本不工作
我无法登录到当前在本地服务器上运行的web应用程序。请帮助,下面是PHP脚本…我们将感谢您的帮助。多谢各位过程php和sql登录脚本不工作,php,mysql,mysqli,Php,Mysql,Mysqli,我无法登录到当前在本地服务器上运行的web应用程序。请帮助,下面是PHP脚本…我们将感谢您的帮助。多谢各位 <?php session_start(); if (isset($_POST['submit'])){ include 'db.inc.php'; //Escape special characters/ $email = mysqli_real_escape_string($conn, $_POST['email
<?php
session_start();
if (isset($_POST['submit'])){
include 'db.inc.php';
//Escape special characters/
$email = mysqli_real_escape_string($conn, $_POST['email']);
$pwd = mysqli_real_escape_string($conn, $_POST['pwd']);
$login = "SELECT * FROM users WHERE user_email='$email'";
$result = mysqli_query($conn, $login);
$resultCheck = mysqli_num_rows($result);
if ($resultCheck < 1) {
header('Location: ../register.php?login=error-zero');
} else {
if ($row = mysqli_fetch_assoc($result)) {
$hashedpwdCheck = password_verify($pwd, $row['user_pwd']);
if ($hashedpwdCheck == false) {
header('Location: ../register.php?login=error'); //IT STOPS RUNNING HERE
exit();
} elseif ($hashedpwdCheck == true){
$_SESSION['name'] = $row['user_name'];
$_SESSION['lastname'] = $row['user_last_name'];
$_SESSION['idnumber'] = $row['user_id_number'];
$_SESSION['cellnumber'] = $row['user_cell'];
$_SESSION['email'] = $row['user_email'];
header('Location: ../inc/profile.php');
exit();
}
}
}
}
您不能从数据库中获取数据,因此您可以使用函数fetch_assoc()。
函数fetch_assoc()将所有结果放入关联数组中
请添加此行$row=$result->fetch_assoc()在您的代码中。如下所示
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.
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$login = "SELECT * FROM users WHERE user_email='$email'";
$result = mysqli_query($conn, $login);
$resultCheck = mysqli_num_rows($result);
$row = $result->fetch_assoc();
.
.
.
确切的错误是?分析错误:语法错误,意外的“;”在第24行的/var/www/html/moreki/inc/login.inc.php中,这是第24行$_会话['name']=$row['user_name'];我的眼睛和IDE都没有发现任何语法错误,你确定这是正确的代码吗?非常确定,也很困惑。。。。。或者逻辑是错误的???您是否尝试过使用“exit();”在“elseif”声明之前退出。这是正确的答案。为什么否决投票???我在本地主机上模拟了这个代码。只有他有这个错误。
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$login = "SELECT * FROM users WHERE user_email='$email'";
$result = mysqli_query($conn, $login);
$resultCheck = mysqli_num_rows($result);
$row = $result->fetch_assoc();
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