[英]Preventing bootstrap modal from disappearing in case validation failure using PHP after submit button is clicked
我的問題是,當用戶單擊“注冊”按鈕時,如何兩次停止模態關閉:
對於第一種情況,如果我重新打開模式,可以看到PHP顯示的驗證錯誤消息。
我已經檢查了這個類似的問題,但還沒有弄清楚該如何解決,因此非常感謝您的幫助。 我想完全了解正在發生的事情和我在做什么。
到目前為止,在這里和那里閱讀之后,我注意到可以通過以下方式實現此目的:
有沒有辦法使用下面的PHP代碼並顯示該代碼中的錯誤/成功消息? 還是必須通過JQuery / JS完成?
我的HTML代碼的PHP
<?php
ob_start();
include('header.php');
include_once("db files/db_connect.php");
if(isset($_SESSION['user_id'])) {
header("Location: index.php");
}
$error = false;
if (isset($_POST['signup'])) {
$name = mysqli_real_escape_string($conn, $_POST['name']);
$email = mysqli_real_escape_string($conn, $_POST['email']);
$password = mysqli_real_escape_string($conn, $_POST['password']);
$cpassword = mysqli_real_escape_string($conn, $_POST['cpassword']);
if (!preg_match("/^[a-zA-Z ]+$/",$name)) {
$error = true;
$uname_error = "Name must contain only alphabets and space";
}
if(!filter_var($email,FILTER_VALIDATE_EMAIL)) {
$error = true;
$email_error = "Please Enter Valid Email ID";
}
if(strlen($password) < 6) {
$error = true;
$password_error = "Password must be minimum of 6 characters";
}
if($password != $cpassword) {
$error = true;
$cpassword_error = "Password and Confirm Password doesn't match";
}
if (!$error) {
if(mysqli_query($conn, "INSERT INTO users(user, email, pass) VALUES('" . $name . "', '" . $email . "', '" . md5($password) . "')")) {
$success_message = "Successfully Registered!";
} else {
$error_message = "Error in registering...Please try again later!";
}
}
}
?>
<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="registrationFormLabel" aria-hidden="true">
<div class="modal-dialog modal-dialog-centered" role="document">
<div class="modal-content" >
<div class="modal-header">
<h5 class="modal-title" id="registrationFormLabel">Register</h5>
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<!-- REGISTRATION FORM -->
<div class="container">
<div class="form-row">
<div class="col">
<form onsubmit="return validateForm()" role="form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post" name="signupform">
<fieldset>
<legend>Sign Up</legend>
<div class="form-group">
<label for="name">Name</label>
<input type="text" name="name" placeholder="Enter Full Name" required value="<?php if($error) echo $name; ?>" class="form-control" />
<span class="text-danger"><?php if (isset($uname_error)) echo $uname_error; ?></span>
</div>
<div class="form-group">
<label for="name">Email</label>
<input type="text" name="email" placeholder="Email" required value="<?php if($error) echo $email; ?>" class="form-control" />
<span class="text-danger"><?php if (isset($email_error)) echo $email_error; ?></span>
</div>
<div class="form-group">
<label for="name">Password</label>
<input type="password" name="password" placeholder="Password" required class="form-control" />
<span class="text-danger"><?php if (isset($password_error)) echo $password_error; ?></span>
</div>
<div class="form-group">
<label for="name">Confirm Password</label>
<input type="password" name="cpassword" placeholder="Confirm Password" required class="form-control" />
<span class="text-danger"><?php if (isset($cpassword_error)) echo $cpassword_error; ?></span>
</div>
<div class="form-group text-center">
<input id="modalSubmit" type="submit" name="signup" value="Sign Up" class="btn btn-primary" formnovalidate />
</div>
</fieldset>
</form>
<span class="text-success"><?php if (isset($success_message)) { echo $success_message; } ?></span>
<span class="text-danger"><?php if (isset($error_message)) { echo $error_message; } ?></span>
</div><!-- / col -->
</div><!-- / form-row -->
<!-- already registered row -->
<div class="row">
<div class="col text-center">
Already Registered? <a href="login.php">Login Here</a>
</div>
</div> <!-- / already registered row -->
</div><!-- / REGISTRATION FORM container-->
</div><!-- / Modal body div -->
</div>
</div>
</div><!-- / Modal -->
當用戶單擊索引頁面上的按鈕時,我的模式打開,這是JQuery代碼
$(document).ready(function(){
$("#myBtn").click(function(){
$("#myModal").modal();
});
});
在文檔末尾( / html ) 腳本標記之前添加以下PHP腳本
<?php
if ($error) {
echo '<script>$("#myModal").modal("show");</script>';
} else {
echo '<script>$("#myModal").modal("hide");</script>';
}
?>
無效:如果輸入無效,則可以停止提交按鈕。
要停止提交,請使用代碼
$(':input[type="submit"]').prop('disabled', true);
顯示模態
$("#myModal").modal('show');
要再次啟用提交,當模態由於錯誤而關閉時,請使用代碼(您將要再次啟用它,以便用戶可以重試)
$("#myModal").on('hidden.bs.modal', function (e) {
$(':input[type="submit"]').prop('disabled', false);
});
為了進行驗證,我將以這種方式進行操作(添加成功變量):
if (!$error) {
if(mysqli_query($conn, "INSERT INTO users(user, email, pass) VALUES('" . $name . "', '" . $email . "', '" . md5($password) . "')")) {
$success = 1;
} else {
$success = 2;
}
然后這樣稱呼它:
<?php if($success == 1){ ?>
<script>
$(document).ready(function() {
$("#yoursuccessmodal").modal('show');
});
</script> <?php } ?>
然后,您可以選擇為其添加淡入淡出,或者讓用戶單擊它。 然后,該模式將在提交提交后顯示。
這是一項更好的工作,因為我在項目中也以這種方式使用了它,並且對我來說就像是一種魅力。
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