Question

我的文件头中包含一个连接文件。标头包含一个会话开始，并且我检查了跨页面的会话ID是否相同。我试图在某些HTML中回显$ _SESSION ['userFirstName']以显示用户名。我不知道为什么它是空白的。除“注意：未定义的索引：userFirstName

”外，Chrome没有其他错误消息

这是我的connect.php

<?php
/*Login handled here*/
$servername = "localhost";
$usernameDB = "root";
$passwordDB = "";
$nameDB = "teacheasy";

//set user and password to values from form
if ( isset($_POST['username']) && isset($_POST['password']) ) { 
    $user = $_POST['username'];
    $pass = $_POST['password'];
} else {
    echo "The values weren't sent";
}

//connect to the database
$_SESSION['connection'] = new mysqli($servername, $usernameDB, $passwordDB,$nameDB);

//Check if the connection was successful
if($_SESSION['connection']->connect_error){
    die("Connection to the database failed: " . $_SESSION['connection']->connect_error);
} else{
//once the DB is connected, get information from the DB to check the records against the data entered
    $sqlUser = "SELECT `teacher_username` FROM `teacher` WHERE `teacher_username`='$user'";
    $sqlPass = "SELECT `password` FROM `teacher` WHERE `password`='$pass'";
    $resultUser = mysqli_query($_SESSION['connection'], $sqlUser);
    $resultPass = mysqli_query($_SESSION['connection'], $sqlPass);
    $textUser = $resultUser->fetch_assoc();
    $textPass = $resultPass->fetch_assoc();

    //get first name and last name to populate the user
    $sqlUserFirstName = "SELECT `first_name` FROM `teacher` WHERE `teacher_username`='$user'";
    $sqlUserLastName = "SELECT `last_name` FROM `teacher` WHERE `teacher_username`='$user'";
    $resultUserFirstName = mysqli_query($_SESSION['connection'], $sqlUserFirstName);
    $resultUserLastName = mysqli_query($_SESSION['connection'], $sqlUserLastName);
    $_SESSION['userFirstName'] = $_POST[$resultUserFirstName->fetch_assoc()];
    $_SESSION['userLastName'] = $_POST[$resultUserLastName->fetch_assoc()];

    //check if the user and password match records in the database
    if($user == $textUser['teacher_username'] && $pass == $textPass['password']){
        //open the calendar if they match
        echo "<script> window.location.assign('../calendar.php'); </script>";
    } else{
        //set this up to load a log in failed page rather than a blank page with error message
        echo "The data entered has no match.";
    }

}

Answer 1

这就是你所做的

$user = $_POST['username']
// "SELECT `first_name` FROM `teacher` WHERE `teacher_username`='$user'" // SQL injection here
$_SESSION['userFirstName'] = $_POST[$resultUserFirstName->fetch_assoc()];

正如@jeroen在评论中所说，$_SESSION['userFirstName']必须为空，因为在$ _POST中没有等于$resultUserFirstName->fetch_assoc()的键返回数组！。您应该收到未定义的索引错误。

$_POST是一个数组，其中包含与http请求一起发布到服务器的变量。除非$_POST['username'] === teacher.first_name和teacher.first_name === teacher.teacher_username

，否则它与数据库查询返回的数据无关。

尝试

$_SESSION['userFirstName'] = $resultUserFirstName->fetch_assoc()['first_name'];

代替

$_SESSION['userFirstName'] = $_POST[$resultUserFirstName->fetch_assoc()];

您还容易受到SQL injection攻击，因此应该始终使用已准备好的语句成为一种感觉。如果您习惯于串联，请查看有关如何switch to prepared statements的答案。

PHP Session Var不跨页面传输

1 个答案: