当我点击按钮打开一个模态弹出窗口时,我有一个名为no of user的按钮..
所以我想要做的是在模式弹出窗口我尝试插入表单,一个字段输入数字..当他们点击提交我想将这些值存储在数据库中
这是我的按钮代码:
<button type="button" class="btn btn-success"style="float:right; margin:3px 0px 3px 3px" id="btnShow">No Of Users</button>
这是我的模态dailog代码:
<script type="text/javascript">
$(function () {
$("#btnShow").click(function(){
$('#demoModal').modal('show');
});
});
<div style="text-align:center; margin-top:10%"></div>
<div class="modal fade" id="demoModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">Bootstrap Modal Popup</h4>
</div>
<div class="modal-body">Hi, Welcome to Aspdotnet-Suresh.com</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="button" class="btn btn-primary">Save changes</button>
</div>
</div>
</div>
任何人都可以帮我怎么做
提前致谢..
答案 0 :(得分:5)
首先,您必须在 index.php 文件中创建模态弹出窗体和ajax脚本
- Modal PopUp表单代码 -
<div style="text-align:center; margin-top:10%"></div>
<div class="modal fade" id="demoModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">Bootstrap Modal Popup</h4>
</div>
<div class="modal-body">
<form method="post">
<input type="text" name="name1" id="name1" placeholder="name">
<br>
<input type="text" name="age1" id="age1" placeholder="age">
<br>
<div id="myCont"></div>
<br>
<input type="submit" id="btn" class="btn" value="SUBMIT">
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="button" class="btn btn-primary">Save changes</button>
</div>
</div>
- 脚本代码 -
<script type="text/javascript" src="js/jquery-3.1.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function (){
$("#btn").click(function (e){
e.preventDefault();
$.ajax({
type: "POST",
async: false,
url:"page1.php",
dataType: 'json',
cache: false,
success: function (result) {
$("#myCont").html(result);
},
error: function () {
alert("server error");
}
});
});
});
</script>
=============================================== =========================
现在创建另一个名为 Page1.php 的页面,其中包含您的数据库插入代码。
<?php
if(isset($_POST))
{
$name = $_POST['name1'];
$age = $_POST['age1'];
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO user_details (name, age)
VALUES ($name, $age)";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
}
?>
我之前已经尝试过,并且工作得非常好。
希望这对你有用。