如何解决如何避免在出现错误的php输入警报后避免页面刷新
<!DOCTYPE html>
<html lang="en-US">
<head>
<title>Captcha Implementation</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width,initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.0/js/bootstrap.min.js"></script>
</head>
<body>
<div class="container" style="width: 600px">
<br/>
<br/>
<div class="panel panel-default">
<div class="panel-body">
<div id="message"></div>
<h4 align="center">Submit the word you see below:</h4>
<p id="captImg" align="center">
<img src="image.php" id="captcha_image"/>
</p>
<!-- <p align="center">Can't read the image? click <a href="javascript:void(0);" class="refreshCaptcha">here</a> to refresh.</p> -->
<form action="index.php" method="post" id="captch_form" name="captch_form">
<div class="form-group">
<label>Enter the Captcha</label>
<input type="text" name="captcha_word" id="captcha_word" class="form-control" placeholder="" value="" autocomplete="off"/>
<div id="message" align="center" class='text-danger'></div>
</div>
<div class="form-group">
<label>Contact Number</label>
<input type="text" name="contact_number" id="contact_number" class="form-control" value="" autocomplete="off"/>
</div>
<div class="form-group" align="center">
<input type="submit" name="submit" id="submit" class="btn btn-info" value="SUBMIT" onclick="captcha_submit()"/>
</div>
</form>
</div>
</div>
</div>
</body>
</html>
<script>
function captcha_submit()
{
var Data = $("#captch_form").serializeArray();
console.log(Data);
$.ajax(
{
url : "action.php?name=captcha_save",type: "POST",data : Data,success:function(data)
{
$('#message').html(data);
}
});
$("#message").slideDown("slow");
e.preventDefault();
}
</script>
<?php
$host = "localhost";
$database = "pmcrb_ambabajaj";
$user = "root";
$password= "";
$db = $database;
$con = mysqli_connect($host,$user,$password,$db);
$Action = @$_GET['name'];
if($Action == 'captcha_save')
{
$captcha_word = @$_POST['captcha_word'];
$contact_number = @$_POST['contact_number'];
$created_date = time();
if($captcha_word == '')
{?>
<div class="alert alert-danger alert-dismissible">
<a href="#" class="close" data-dismiss="alert" aria-label="close">×</a>
Please Enter Captcha.
</div>
<?php }
/*else if($captcha_word == $_SESSION['captcha_code'])
{?>
<div class="alert alert-danger alert-dismissible" role="alert">
<button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>
<strong style="color:#fff;text-align:center;font-size:14px;font-family:Times New Roman,Times,serif;margin-left:50px;">Warning!</strong> Invalid Captcha Code
</div>
<?php }*/
else if($contact_number == '')
{?>
<div class="alert alert-danger alert-dismissible" role="alert">
<button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>
<strong style="color:#fff;text-align:center;font-size:14px;font-family:Times New Roman,serif;margin-left:50px;">Warning!</strong> Please Enter Contact Number
</div>
<?php }
else
{
$sql = "insert into pm1captcha(`created_date`,`contact_number`,`captcha_word`)values('$created_date','$contact_number','$captcha_word')";
$qry = mysqli_query($con,$sql);
?>
<div class="alert alert-success alert-dismissible" role="alert">
<button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">×</span></button>
<strong style="color:#fff;text-align:center;font-size:14px;font-family:Times New Roman,serif;margin-left:50px;">Success!</strong> Data Saved Successfully
</div>
<?php }
}
?>
我已经编写了一些代码来将验证码详细信息保存到数据库表中,数据已正确保存,但是如果文件字段为空,则页面每次都将刷新。如何避免每次在php中出现警报后刷新页面。谁能帮帮我吗。在这里,我在index.php页面中编写了html代码,而在action.php页面中编写了php代码
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