我有一个问题,一旦我激活或停用我的员工,数据表搜索和排序将消失。我认为我的错误是在加载部分。使用div id加载特定页面的正确方法是什么?
时的输出因为我所做的是在我的base_url +“adminpage / userpage”中,它是用户和员工的一页。我刚刚做的是当我按下查看人员时它将隐藏用户并将显示在同一链接中的员工页面。
我的代码在这里
$(document).ready(function(){
$("#user-page").show();
$("#viewstaff").on("click",function(){
$("#user-page").hide();
$('#staff-page').show();
});
$("#viewuser").on("click",function(){
$("#staff-page").hide();
$("#user-page").show();
});})
我的激活代码在这里
function activate_staff(id){
swal({
title: 'Are you sure?',
text: "You want to activate this user?",
type: 'info',
showCancelButton: true,
confirmButtonColor: '#3085d6',
cancelButtonColor: '#d33',
confirmButtonText: 'Yes, Deactivate it!'
}).then(function() {
$.ajax({
url : base_url+"adminpage/staff_deactivate/"+id,
type: "POST",
success: function(data){
$('#dataTables-example1').DataTable();
var result = JSON.parse(data);
if(result===1){
swal({
title: 'Deactivate Success',
// text: "will remove within 1 sec",
type: 'success',
timer: 1500,
showConfirmButton: false
}).done()
window.setTimeout(function(){
$("#wrapper-staff").load(base_url+"adminpage/useraccount #wrapper-staff > *");
} ,1500)
}
},
error: function (jqXHR, textStatus, errorThrown){
swal("Oops...", "Something went wrong :(", "error");
}
});
})}
和停用是
function deactivate_staff(id){
swal({
title: 'Are you sure?',
text: "You want to deactivate this user?",
type: 'info',
showCancelButton: true,
confirmButtonColor: '#3085d6',
cancelButtonColor: '#d33',
confirmButtonText: 'Yes, Activate it!'
}).then(function() {
$.ajax({
url : base_url+"adminpage/staff_activate/"+id,
data: $(this).serialize(),
type: "POST",
success: function(data){
var result = JSON.parse(data);
if(result===1){
swal({
title: 'Activated Success',
// text: "will remove within 1 sec",
type: 'success',
timer: 1500,
showConfirmButton: false
}).done()
window.setTimeout(function(){
$("#wrapper-staff").load(base_url+"adminpage/useraccount #wrapper-staff > *");
} ,1500)
}
},
error: function (jqXHR, textStatus, errorThrown){
swal("Oops...", "Something went wrong :(", "error");
}
});
}) }
答案 0 :(得分:2)
您必须将此DataTable更改为服务器端,如example。之后,您可以在每行上添加按钮,如下所示:
table = $('#dataTables-example').DataTable({
"serverSide": true, //Feature control DataTables' server-side processing mode.
// Load data for the table's content from an Ajax source
"ajax": {
"url": "<?php echo site_url('serversidedt/ajax_dt')?>",
"type": "POST"
},
"columns": [
{data: "fname"},
{data: "lname"},
{data: "email"},
{data: "status"},
{data: myButton, searchable: false, orderable: false}
]
});
function myButton(data, type, dataToSet) {
if(dataToSet.status == 1){
return "<button class='btn btn-primary'>Active</button>";
}else{
return "<button class='btn btn-primary'>Desactive</button>";
}
};
你必须像这样刷新你的DataTable:
table.ajax.reload( null, false )
希望有所帮助。