Question

我创建了一个带有显示jQuery的图片幻灯片的页面。我还搜索到用户可以从数据库中搜索房屋，并且此代码可以查找。当我添加php代码以便允许用户登录并尝试运行它时页面出现空白，为什么会这样？

这是我的代码

session_start();

  include "connect.php";

  if (isset($_POST['username']) and isset($_POST['password'])){

$username = $_POST['username'];

$password = $_POST['password'];

$query = ($con "SELECT * FROM login WHERE username='$username' and password= '$password'");

$result = mysqli_query($query) or die(mysqli_error());

$count = mysqli_num_rows($result);

if ($count == 1){

$_SESSION['username'] = $username;

}else {

echo "Invalid Login Credentials.";

}


if (isset($_SESSION['username'])){

$username = $_SESSION['username'];

echo "Hello " . $username . "";
echo "This is the Members Area";

echo "<a href='logout.php'>Logout</a>";

}
?>

Answer 1

首先，您没有将您的连接传递给您的查询，并且您有一个缺失的大括号。

if (isset($_POST['username']) and isset($_POST['password']))的那个应该封装整个PHP。

旁注：使用$con作为连接变量的参数。

<?php
session_start();

  include "connect.php";

  if (isset($_POST['username']) and isset($_POST['password'])){

$username = $_POST['username'];

$password = $_POST['password'];

$query = "SELECT * FROM login WHERE username='$username' and password='$password'";

$result = mysqli_query($con, $query) or die(mysqli_error());

$count = mysqli_num_rows($result);

if ($count == 1){

$_SESSION['username'] = $username;

}else {

echo "Invalid Login Credentials.";

}


if (isset($_SESSION['username'])){

$username = $_SESSION['username'];

echo "Hello " . $username . "";
echo "This is the Members Area";

}

} // closing brace for if (isset($_POST['username']) and isset($_POST['password']))

echo "<a href='logout.php'>Logout</a>";

?>

我注意到您可能以纯文本格式存储密码。如果是这种情况，则非常气馁。

我建议您使用CRYPT_BLOWFISH或PHP 5.5＆＃39 {s} password_hash()功能。对于PHP＆lt; 5.5使用password_hash() compatibility pack。

当我使用mamp运行我的PHP代码时什么都没有显示

1 个答案: