如何在php中使用bootstrap发生错误时显示错误

我正在使用 php 和 bootstrap 构建一个联系表单,我想在发生错误时显示 bootstrap 警报。所以我现在的问题是我已经让代码工作并且它检查错误但是在错误发生之前显示错误警报对话框


我试图将错误对话框移到 HTML 下方,但我仍然遇到相同的错误


$email = $name = $subject = $message = '';


$errors = ['email' => '', 'name' => '', 'subject' => '', 'message' => ''];


if (isset($_POST['submit'])) {


    // check email

    if (empty($_POST['email'])) {

        $errors['email'] = "An email is required !";

    } else {

        $email = $_POST['email'];

        if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {

            $errors['email'] = "Enter a validemail address !";

        }

    } 



<div class="form-group">


    <label for="email" class="bmd-label-floating">Email address</label>

    <input type="email" class="form-control" name="email" value="<?php htmlspecialchars($email)?>">

    <div class="mt-2 alert alert-warning alert-dismissible fade show" role="alert">

        <?php echo $errors['email'] ?>

    </div>

</div>

预期的行为应该是在错误发生时在错误对话框中显示错误,而不是在错误发生之前出现错误对话框


忽然笑
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2回答

慕桂英3389331

您想用 an 包装模态if以检查是否确实存在错误。<?php if($errors['email']): ?><div class="mt-2 alert alert-warning alert-dismissible fade show" role="alert">&nbsp; &nbsp; <?php echo $errors['email'] ?></div><?php endif; ?>

守着星空守着你

在显示错误之前添加检查。喜欢下面<?phpif(isset($errors['email']) && $errors['email']!=''){&nbsp;?><div class="mt-2 alert alert-warning alert-dismissible fade show" role="alert">&nbsp; &nbsp; &nbsp; &nbsp; <?php echo $errors['email']; ?></div><?php } ?>
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