Question

我需要使用PHP登录脚本并编写该脚本（请参见下面的脚本），但是表单没有通过URL（使用GET或POST）将输入文件中的变量发送到同一文件中

<?php
session_start();
$admin=true;

function controle($uname, $pword, $admin){

$datatable = "my_table";
$servername = 'localhost';
$username = 'admin';
$password = '1234';
$database = 'myDB';

//Create connection
$con = mysqli_connect($servername, $username, 
$password, $database);

if ($con == false){
die("STATUS: Error: ".mysqli_error_connect());
}

if ($admin == true){
    $sql = "SELECT * FROM ".$datatable; 
}
$result = mysqli_query($con,$sql);
$rows=mysqli_fetch_assoc($result);

$sql2 = "SELECT COUNT(*) AS total FROM ".$datatable;
$result = mysqli_query($con,$sql2);

$row2 = $result->fetch_row();
$total_records = $row2[0];

if ($total_records > 0){

    $gebruikersnaam = $rows['username'];
    $wachtwoord = $rows['password'];
    if ($pword != $wachtwoord){
        return false;
    } else {
        return $gebruikersnaam;
    }
} else {
    return false;
}
}

// Check name and password
if (isset($_POST['verzonden'])){
$username = $_POST['username'];
$password = md5($_POST['wachtwoord']);
$login_ok = controle($username, $password, $admin);
if (login_ok != false){
    //correct
    $_SESSION['username'] = $username;
    $_SESSION['password'] = $password;
}
}
if (controle($_SESSION['username'], 
$_SESSION['wachtwoord'], $admin) == false) {
echo "<form method='post' 
action='".$_SERVER['PHP_SELF']."?";
reset($_GET); // put the array pointer to 0 when 
starting
// Send the variables again
while($getvar = each($_GET)){
    $varName = $getvar['key'];
    $varValue = $getvar['value'];
    echo "$varName=$varValue&";
}

echo "'><br><br>";
echo "Name: ";
echo "<input type='text' name='username'>";
echo "<br>";
echo "Password: ";
echo "<input type='text' name='wachtwoord'>";
echo "<br>";
echo "<input type='submit' value='log in' 
name='verzonden'>";
echo "</form>";
if ($admin == "true"){
    echo "<p>-- ADMIN status is vereist!";
}
exit;
}
?>

但是该表格似乎没有发送任何内容。.我期待的是：

authentication.php?username=MYNAME&wachtwoord=4321

与数据库的连接正常，我可以从那里读取变量。我没有ID，为什么我的表单不发送仅获得“ authentication.php”的变量？

Answer 1

您正在混淆$ _GET和$ _POST。您的表单正在使用$ _POST，但是您正在寻找$ _GET变量。

我已更改您的代码，以使其正常工作。如果您想调试并查看整个过程中发生的事情，我留了一些var_dump行

我已将您的数据库连接设置为全局连接，以便您仅连接一次，并且您可以在需要时再次访问它。我已将您的GET更改为POST。当您登录或何时需要登录时，我已经显示了它。我也为您建立了一个注销部分。

<?php
session_start();
$admin=true;

$datatable = "my_table";
$servername = 'localhost';
$username = 'nick';
$password = 'nickd18';
$database = 'test';

// show session variables for debuggin;
var_dump($_SESSION);

// you may want to destroy session variables at some point when debugging (should be a log out but we can use this line for now)
// 
if (isset($_GET) && $_GET['logout'] == '1') {
    session_destroy();
    $login = "http://".$_SERVER[HTTP_HOST].$_SERVER['PHP_SELF'];
    die(header('Location: '.$login));
}


//Create connection
global $con;
$con = mysqli_connect($servername, $username, $password, $database);

if ($con == false){
    die("STATUS: Error: ".mysqli_error_connect());
}


function check_login($uname, $pword, $admin){
    global $con;
    $sql = "SELECT * FROM `my_table` WHERE `username` = '$uname' AND `password` = '$pword' LIMIT 1";
    //var_dump($sql);
    $result = mysqli_query($con,$sql);
    //var_dump($result);
    $rows=mysqli_fetch_assoc($result);
    if ($rows) {
        /* successful login */
        return true;
    }  else {
        /* failed login */
        return false;
    }


}

// Check name and password
if (isset($_POST['verzonden'])){
    //var_dump($_POST);
    $username = $_POST['username'];
    $password = md5($_POST['wachtwoord']);
    $login_ok = check_login($username, $password, $admin);
    if ($login_ok != false){
        //correct
        $_SESSION['username'] = $username;
        $_SESSION['password'] = $password;
    }
}
if (!$_SESSION['username']) {
    echo "<h2>You are not logged in</h2>";
    echo "<form method='post' action='".$_SERVER['PHP_SELF']."?";

    echo "'><br><br>";
    echo "Name: ";
    echo "<input type='text' name='username'>";
    echo "<br>";
    echo "Password: ";
    echo "<input type='text' name='wachtwoord'>";
    echo "<br>";
    echo "<input type='submit' value='log in' 
    name='verzonden'>";
    echo "</form>";
    if ($admin == "true"){
        echo "<p>-- ADMIN status is vereist!";
    }
    exit;
} else {
    echo "<h1>You are logged in as ".$_SESSION['username']."</h1>";
    echo "<p><a href='?logout=1'>Log out</a>";
}
?>

PHP脚本不会从表单发送变量

1 个答案: