我创建了一个表作为company_list。该表包含以下字段
1)公司名称2)国家/地区3)垂直
所以,假设有以下5家公司
公司名称:水,地球,太阳,月亮,星星
国家/地区:水=英国,地球=英国,太阳=澳大利亚,月亮,=印度,星星=德国
垂直:水,地球,太阳=世界自然基金会和月亮,星星=社会福利
第一个HTML是一个筛选选项
第一国第二纵队
因此,如果我选择了英国国家/地区,垂直行业应自动选择WWF(不包括澳大利亚国家/地区的公司名称)
现在,发布此信息后,我将具有一个按钮以首先获取符合上述条件的单个公司详细信息。
我设计的代码不是添加过滤器,而是会提取所有公司。
请对此提供帮助。
1)索引页
<?php
$conn = mysqli_connect("localhost","root","","accounts");
$sql = "SELECT * FROM company_list";
$res= mysqli_query($conn,$sql);
$res1=mysqli_query($conn,$sql);
?>
<!DOCTYPE html>
<html lang="en" dir="ltr">
<head>
<meta charset="utf-8">
<title></title>
<!-- Bootstrap 3.3.7 -->
<link rel="stylesheet" href="bower_components/bootstrap/dist/css/bootstrap.min.css">
<!-- Font Awesome -->
<link rel="stylesheet" href="bower_components/font-awesome/css/font-awesome.min.css">
<!-- Ionicons -->
<link rel="stylesheet" href="bower_components/Ionicons/css/ionicons.min.css">
<!-- DataTables -->
<link rel="stylesheet" href="bower_components/datatables.net-bs/css/dataTables.bootstrap.min.css">
<!-- Theme style -->
<link rel="stylesheet" href="dist/css/AdminLTE.min.css">
<!-- AdminLTE Skins. Choose a skin from the css/skins
folder instead of downloading all of them to reduce the load. -->
<link rel="stylesheet" href="dist/css/skins/_all-skins.min.css">
<style media="screen">
.navbar-inverse {
background-color: #4a8cbb;
border-color: #4a8cbb;
}
#row1{
margin-left: 100px;
}
</style>
</head>
<body>
<nav class="navbar navbar-inverse">
<div class="container-fluid">
<div class="navbar-header">
<select class="form-control">
<?php while ($row=mysqli_fetch_array($res)):; ?>
<option value="">SELECT COUNTRY</option>
<option value=""><?php echo $row['country']; ?></option>
<?php endwhile; ?>
</select>
</div>
<div class="navbar-header" id="row1">
<select class="form-control">
<?php while ($row1=mysqli_fetch_array($res1)):; ?>
<option value="">SELECT VERTICAL</option>
<option value=""><?php echo $row1['vertical']; ?></option>
<?php endwhile; ?>
</select>
</div>
<div class="navbar-header" id="row1">
<button id="fetch_company" class="btn btn-success btn-lg btn-block btn-huge">FETCH COMPANY DETAILS</button>
</div>
</div>
</nav>
<div class="row">
<div class="col-md-3">
Company Name:<span id="com_get_data"></span>
</div>
</div>
</body>
<script type="text/javascript" src="bower_components/jquery/dist/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
var count = 0;
$("#fetch_company").click(function(){
//alert("Working");
count = count + 1;
$("#com_get_data").load("get_data.php",{countNew:count});
});
});
</script>
</html>
2)Get_data.php
<?php
$conn = mysqli_connect("localhost","root","","accounts");
$countNew = $_POST['countNew'];
$sql = "SELECT * FROM company_list LIMIT $countNew";
$res = mysqli_query($conn,$sql);
if (mysqli_num_rows($res) > 0) {
while ($row = mysqli_fetch_array($res)) {
echo $row['company_name'];
}
}
?>
答案 0 :(得分:0)
请使用$.ajax
jQuery代码
<script type="text/javascript">
$(document).ready(function(){
var count = 0;
$("#fetch_company").click(function(){
//alert("Working");
count = count + 1;
$.ajax({
type:"POST",
url: "get_data.php",
data:{countNew:count},
cache: false,
success: function(result){
$("#com_get_data").html(result);
}
});
});
});
</script>
Php代码
<?php
$conn = mysqli_connect("localhost","root","","accounts");
$countNew = $_POST['countNew'];
$sql = "SELECT * FROM company_list LIMIT $countNew";
$res = mysqli_query($conn,$sql);
if (mysqli_num_rows($res) > 0) {
while ($row = mysqli_fetch_array($res)) {
echo $row['company_name'];
}
}
else
{
echo 'No result found';
}
?>
注意:在这里,您需要循环访问槽中的数据,然后在string array
中返回数据,然后在JSON
中添加数据
请在此处查看更多详细信息Click Here