我是PHP初學者。 我的代碼在php錯誤中更改innerHTML
<?php
session_start();
$username = "ADMIN";
$host = "localhost";
$password = "chmuhammadsohaib123";
$database = "USER";
$con = mysqli_connect($host, $username, $password, $database);
$USERNAME = $_POST["lusername"];
$PASSWORD = $_POST["lpassword"];
if (isset($_POST["login"])) {
if (isset($_POST["loggedin"])) {
setcookie("RAUSERNAME", $USERNAME);
setcookie("RAPASSWORD", $PASSWORD);
}
$_SESSION["SRAUSERNAME"] = $USERNAME;
$_SESSION["SRAPASSWORD"] = $PASSWORD;
}
if (isset($_POST["login"])) {
$data = mysqli_query($con, "SELECT * FROM `INFO` WHERE `USERNAME` = '$USERNAME'");
if (mysqli_num_rows($data)>0) {
echo "<script type='text/javascript'>window.location.replace('../');</script>";
}
else {
print("<script type='text/javascript'>document.getElementsByClassName('errors').innerHTML = '<h1 class='redback'>SORRY, BUT THIS ACCOUNT DOESN'T EXISTS</h1>';</script>");
}
}
?>
我的HTML頁面
<body>
<div class="errors"></div>
<fieldset class="replacement">
<legend>LOGIN</legend>
<h1>LOGIN WITH YOUR INFORMATION</h1><br><br>
<form method="POST" action="<?php $_SERVER["php_self"]; ?>">
<input type="text" name="lusername" placeholder="YOUR USERNAME">
<input type="password" name="lpassword" placeholder="YOUR PASSWORD" class="password">
<br>
<br>
<label>KEEP ME LOGGED IN: </label>
<input type="checkbox" name="loggedin" checked>
<br><br>
<input type="submit" name="login" value="LOGIN"></form>
</fieldset>
</div>
</body>
</html>
如上所述當我改變錯誤的innerHTML,它不會改變。它說;中缺少控制檯或有時是錯誤是空。我該如何解決它?
怎樣的PHP涉及到HTML網頁? – chris85
你對SQL注入開放,並有錯誤的登錄邏輯,你需要檢查密碼(應該被哈希)。 – chris85
通過在php中請求 –