[英]show success message after form submit
我想在更新语句时显示一条消息但是当我这样做时按钮类型是提交它刷新页面并且当我将按钮类型更改为按钮时该消息不会显示它显示消息但不执行更新操作这是我的代码html 代码
<div class="alert alert-success hidden" id="success-alert">
<span class="glyphicon glyphicon-ok"></span> User Name Changed Succsefully
</div>
<div class="row">
<div class="col-md-6">
New username: <input type="text" class="form-control" name="user" placeholder="New username">
</div>
</div>
<div class="col-auto my-1">
<button type="submit" name="changeuser" id="user" class="btn btn-primary">change username</button>
</div>
</fieldset>
</form>
代码
<script type="text/javascript">
$(document).ready(function() {
$("#user").click(function() {
$('#success-alert').removeClass('hidden');
$('#success-alert').delay(5000).fadeOut('slow');
});
$("button.close").click(function() {
$('#success-alert').addClass('hidden');
});
});
代码
if (isset($_POST['changeuser']))
{
$username=$_POST['user'];
$sql="SELECT * from user_account where username='$username'";
$result=mysqli_query($conn,$sql);
if(mysqli_num_rows($result)==1)
{
echo '<script language="javascript">';
echo 'alert("User name already exits Please Chose Another One")';
echo '</script>';
}
else
{
$up = mysqli_query($conn, "UPDATE user_account SET username='$username' WHERE id='$id'");
}
}
将您的 php 代码更改为:
$up = '';
if (isset($_POST['changeuser']))
{
$username=$_POST['user'];
$sql="SELECT * from user_account where username='$username'";
$result=mysqli_query($conn,$sql);
if(mysqli_num_rows($result)==1)
{
echo '<script language="javascript">';
echo 'alert("User name already exits Please Chose Another One")';
echo '</script>';
}
else
{
$up = mysqli_query($conn, "UPDATE user_account SET username='$username' WHERE id='$id'");
}
}
并像这样更改您的 html 部分:
<?php
if($up) {
?>
<div class="alert alert-success" id="success-alert">
<span class="glyphicon glyphicon-ok"></span> User Name Changed Succsefully
</div>
<?php
}
?>
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