Question

我是PHP的初学者。我的代码

    <?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>");
        }
    }
?>

MY HTML PAGE

<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，当我更改错误的innerHTML时，它不会发生变化。它表示控制台中缺少 ; ，有时错误 null 。我该如何解决？

Answer 1

在您回显javascript代码时，dom中不存在标识为errors的html元素。因此，getElementById的返回将始终未定义。

<script>document.getElementById("errors")...</script>
... some more html
<div id="errors"></div>

您可以通过在 dom文档准备好之后调用javascript代码来解决此问题。使用jQuery，你可以这样做

// event handler for document ready $(function() { // at this point, the dom is ready and the 'errors' id exists $('#errors').html("some error message"); });

这可行，但似乎有点不必要。更好的方法是使用php回显实际的错误消息，不要使用javascript来执行此操作。

$error = false; if (mysqli_num_rows($data)>0) { header('location: ../'); } else { $error = '<h1 class="redback">SORRY, BUT THIS ACCOUNT DOESN\'T EXISTS</h1>'; }

以后

<div class="errors"> <?php if ($error) echo $error; ?> </div>

在php错误中更改innerHTML

1 个答案: