[英]Contact Form 7: Submit button doesn't work in Twitter Bootstrap Modal
[英]Why is not submit button work on bootstrap modal popup
这是我的 PHP 代码
<?php
include('dbConfig.php');
?>
<?php
if(isset($_POST['uploads']))
{
$brands_name=$_POST['brand_name'];
$product=$_POST['product'];
$model_no=$_POST['model_no'];
/// File Add
$bills_file = $_FILES['billscopy']['name'];
// file Temp
$tembills_file =$_FILES['billscopy']['tmp_name'];
//uploading an image to its folder
move_uploaded_file($tembills_file,"mybills/$bills_file");
mysql_query("INSERT INTO `mybills`(`user_id`, `user_name`, `user_email`, `brands_name`,`product_name`,`model_no`,`bills_copy`)
VALUES ('$user_id','$userName','$userEmail','$brands_name','$product','$model_no','$bills_file')") or die(mysql_error());
echo "<div class='alert alert-success'>Bills Copy Added Successfully.</div>";
}
?>
这是我的模式弹出代码:
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Add Bills </h4>
</div>
<div class="modal-body">
<form method="post" action="#" enctype="multipart/form-data">
<div class="form-group col-md-6 col-sm-6">
<?php
include('dbConfig.php');
$query = mysql_query("SELECT * FROM `brands` ORDER BY brands_name ");
$rowCount = mysql_num_rows($query);
?>
<center><select name="brand_name" id="course" class="input-xxlarge form-control">
<option value="">Select Brands</option>
<?php
if($rowCount > 0){
while($row =mysql_fetch_array($query)){
echo '<option value="'.$row['brands_id'].'">'.$row['brands_name'].'</option>';
}
}else{
echo '<option value="">Brands not available</option>';
}
?>
</select></center>
</div>
<center><div class ="form-group col-md-6 col-sm-6">
<select name="product" id="subject" class="input-xxlarge form-control">
<option value="">Select Product first</option>
</select>
</div></center>
<center><div class ="form-group col-md-6 col-sm-6">
<select name="model_no" id="testname" class="input-xxlarge form-control">
<option value="">Select Model No first</option>
</select>
</div></center>
<center><div class ="form-group col-md-6 col-sm-6">
<input type="file" name="billscopy" class="form-control" accept="image/*" capture="camera" data-name="full-name" required></center>
<br></center>
<!---<center><input class="form-control" id="full-name-field" type="file" name="billscopy" accept="image/*" capture="camera" data-name="full-name" required>
<div class="separator-fields"></div></center>
</div>--->
</div>
<div class="modal-footer">
<button type="submit" class="btn btn-success btn btn-lg" name="uploads">Upload</button>
</div>
</form>
</div>
</div>
当我第一次运行代码时,提交按钮不起作用,但是当刷新页面时,提交按钮完美地工作并且所有动态数据都提交成功。
这是我的直播 URL: http://meraapp.solidaleinfotech.com/ 。 打开此链接并登录
登录 ID=demo@gmail.com
通过=12345
然后登录点击我的账单,然后在点击相机图像模式弹出窗口时点击相机图像并填写所有字段并点击上传按钮所以上传按钮不起作用但刷新页面按钮完美工作和数据插入。
你在错误的地方写了<form method="post" action="#" enctype="multipart/form-data">
。 查看您在何处关闭</form>
以及在何处开始<form>
。 <form>
它应该在模态主体之上。
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