本文介绍了复选框和检查Datatable Jquery PHP Mysql中的所有功能的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
以下是用于从json中的数据库表中将数据提取到Datatable的php脚本;
The below is the php script for fetchingd data from database table in json to Datatable;
<?php
/* Database connection start */
$servername = "localhost";
$username = "root";
$password = "password1";
$dbname = "test";
$conn = mysqli_connect($servername, $username, $password, $dbname) or die("Connection failed: " . mysqli_connect_error());
/* Database connection end */
// storing request (ie, get/post) global array to a variable
$requestData= $_REQUEST;
$columns = array(
// datatable column index => database column name
0 =>'employee_name',
1 => 'employee_salary',
2=> 'employee_age'
);
// getting total number records without any search
$sql = "SELECT employee_name, employee_salary, employee_age ";
$sql.=" FROM employee";
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
$totalData = mysqli_num_rows($query);
$totalFiltered = $totalData; // when there is no search parameter then total number rows = total number filtered rows.
$sql = "SELECT employee_name, employee_salary, employee_age ";
$sql.=" FROM employee WHERE 1=1";
if( !empty($requestData['search']['value']) ) { // if there is a search parameter, $requestData['search']['value'] contains search parameter
$sql.=" AND ( employee_name LIKE '".$requestData['search']['value']."%' ";
$sql.=" OR employee_salary LIKE '".$requestData['search']['value']."%' ";
$sql.=" OR employee_age LIKE '".$requestData['search']['value']."%' )";
}
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
$totalFiltered = mysqli_num_rows($query); // when there is a search parameter then we have to modify total number filtered rows as per search result.
$sql.=" ORDER BY ". $columns[$requestData['order'][0]['column']]." ".$requestData['order'][0]['dir']." LIMIT ".$requestData['start']." ,".$requestData['length']." ";
/* $requestData['order'][0]['column'] contains colmun index, $requestData['order'][0]['dir'] contains order such as asc/desc */
$query=mysqli_query($conn, $sql) or die("employee-grid-data.php: get employees");
$data = array();
while( $row=mysqli_fetch_array($query) ) { // preparing an array
$nestedData=array();
$nestedData[] = $row["employee_name"];
$nestedData[] = $row["employee_salary"];
$nestedData[] = $row["employee_age"];
$data[] = $nestedData;
}
$json_data = array(
"draw" => intval( $requestData['draw'] ), // for every request/draw by clientside , they send a number as a parameter, when they recieve a response/data they first check the draw number, so we are sending same number in draw.
"recordsTotal" => intval( $totalData ), // total number of records
"recordsFiltered" => intval( $totalFiltered ), // total number of records after searching, if there is no searching then totalFiltered = totalData
"data" => $data // total data array
);
echo json_encode($json_data); // send data as json format
?>
下面是我的数据表页面;
And below is my datatable page;
<script type="text/javascript" language="javascript" >
$(document).ready(function() {
var dataTable = $('#employee-grid').DataTable( {
"processing": true,
"serverSide": true,
"ajax":{
url :"employee-grid-data.php", // json datasource
type: "post", // method , by default get
error: function(){ // error handling
$(".employee-grid-error").html("");
$("#employee-grid").append('<tbody class="employee-grid-error"><tr><th colspan="3">No data found in the server</th></tr></tbody>');
$("#employee-grid_processing").css("display","none");
}
}
} );
} );
</script>
我试图在第一栏中添加复选框以选择一行或多行或全部行并提交.
I m trying to add checkbox in first column to select one row or multiple or all row and submit.
但是我不能这样做,有人可以帮我吗?
But I am not able to do so, can anyone help me out to do it?
推荐答案
请参见 jQuery jQuery DataTables的DataTables Checkboxes 插件.
var table = $('#example').DataTable({
'columnDefs': [
{
'targets': 0,
'checkboxes': {
'selectRow': true
}
}
],
'select': {
'style': 'multi'
},
'order': [[1, 'asc']]
});
这篇关于复选框和检查Datatable Jquery PHP Mysql中的所有功能的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!