使用 ajax 通过分页和排序将行动态添加到数据表
Dynamically adding rows to datatable using ajax with pagination and sorting
我正在努力实现 https://almsaeedstudio.com/themes/AdminLTE/pages/tables/data.html - "Data Table With Full Features"
当我静态添加 tbody 时,分页和排序工作正常,但是当我使用 jquery 添加 tbody 时,如下所示,添加了行但分页和排序失败。
HTML
<table id="tblItems">
<thead>
<tr>
<th>code</th>
<th>Name</th>
<th>Description</th>
<th>Image</th>
<th>Parent</th>
<th>Location</th>
<th>Category</th>
</tr>
</thead>
</table>
jquery
$(document).ready(function() {
$('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
$('#tblItems').append('<tbody><tr><td>asdsa34id</td><td> asdsa34id </td><td>asdsa34id </td><td> asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td></tr></tbody>');
});
https://jsfiddle.net/techjerk2013/vwpsxhaL/
更新代码
更新后的代码不会填充 table 尽管有来自响应的数据。虽然我将 deferRender 设置为 true,但数据table 仍然是空的。
$(document).ready(function() {
PopulateItemsTable();
BindTable();
});
function BindTable() {
$("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
function PopulateItemsTable() {
var txt = "";
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
if (jsonObject) {
var len = jsonObject.length;
if (len > 0) {
for (var i = 0; i < len; i++) {
if (jsonObject[i].Id) {
Id = jsonObject[i].Id;
}
else {
Id = '';
}
if (jsonObject[i].Name) {
Name = jsonObject[i].Name;
}
else {
Name = '';
}
if (jsonObject[i].Description) {
Description = jsonObject[i].Description;
}
else {
Description = '';
}
if (jsonObject[i].Image) {
Image = jsonObject[i].Image;
}
else {
Image = '';
}
if (jsonObject[i].Parent) {
Parent = jsonObject[i].Parent;
}
else {
Parent = '';
}
if (jsonObject[i].Location) {
Location = jsonObject[i].Location;
}
else {
Location = '';
}
Category = '';
txt += "<tr><td>" + Id + "</td><td>" + Name + "</td><td>" + Description + "</td><td>" + Image + "</td><td>" + Parent + "</td><td>" + Location + "</td><td>" + Category + "</td></tr>";
$('#tblItems').append('<tbody>' + txt + '</tbody>');
}
}
else {
$("#tblItems").append("No records found");
}
}
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
回答
在下面回答的人的帮助下,下面的代码按预期工作。
<script type="text/javascript">
var myTable;
$(document).ready(function () {
BindItemTable();
PopulateItemsTable();
});
function BindItemTable() {
myTable = $("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
function PopulateItemsTable() {
$.ajax({
type: "POST",
url: "ItemManagement.aspx/SearchIdList",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function (item) {
var result = [];
result.push(item.Id);
result.push(item.Name);
result.push(item.Description);
result.push(item.Image);
result.push(item.Parent);
result.push(item.Location);
result.push("");
return result;
});
myTable.rows.add(result);
myTable.draw();
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
</script>
不要直接将行添加到 table 标记,而是将其添加到 DataTable 实例,然后使用 .draw()
方法。无论如何,添加到 DataTable 实例都会在内部将其添加为 tbody
。像这样的事情应该做
var mytable = $('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
mytable.row.add(['asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id']);
mytable.draw();
这是一个演示 https://jsfiddle.net/dhirajbodicherla/vwpsxhaL/1/
还阅读 how to add rows to DataTable 他们的文档以供进一步参考
更新
您可以使用 rows.add()
(复数)并执行类似这样的操作
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function(item){
var result = [];
result.push(item.Id);
// .... add all the values required
return result;
});
myTable.rows.add(result); // add to DataTable instance
myTable.draw(); // always redraw
var myTable;
$(document).ready(function() {
myTable = $("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
PopulateItemsTable();
});
function PopulateItemsTable() {
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function(item){
var result = [];
result.push(item.Id);
// .... add all the values required
return result;
});
myTable.rows.add(result); // add to DataTable instance
myTable.draw(); // always redraw
}
});
}
如果您正在修改 table html 使用 jQuery 而不是插件提供的 api,那么您必须再次调用插件,以便它会根据修改后的内容重新实例化html.
$(document).ready(function() {
$('#tblItems').append('<tbody><tr><td>asdsa34id</td><td> asdsa34id </td><td>asdsa34id </td><td> asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td></tr></tbody>');
$('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
});
演示 https://jsfiddle.net/8hyr08xb/
根据编辑后的问题更新
试试这个
function PopulateItemsTable() {
var txt = "";
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d), html = [];
if (jsonObject) {
var len = jsonObject.length;
if (len > 0) {
for (var i = 0; i < len; i++) {
if (jsonObject[i].Id) {
Id = jsonObject[i].Id;
}
else {
Id = '';
}
if (jsonObject[i].Name) {
Name = jsonObject[i].Name;
}
else {
Name = '';
}
if (jsonObject[i].Description) {
Description = jsonObject[i].Description;
}
else {
Description = '';
}
if (jsonObject[i].Image) {
Image = jsonObject[i].Image;
}
else {
Image = '';
}
if (jsonObject[i].Parent) {
Parent = jsonObject[i].Parent;
}
else {
Parent = '';
}
if (jsonObject[i].Location) {
Location = jsonObject[i].Location;
}
else {
Location = '';
}
Category = '';
html.push("<tr><td>" + Id + "</td><td>" + Name + "</td><td>" + Description + "</td><td>" + Image + "</td><td>" + Parent + "</td><td>" + Location + "</td><td>" + Category + "</td></tr>");
}
$('#tblItems')
.append('<tbody>' + html.join('') + '</tbody>')
.DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
else {
$("#tblItems").append("No records found");
}
}
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
相信我,我做了以上所有事情 none 无需编写长行代码就可以按预期工作
对我来说,解决方案非常简单快捷。
现在要让数据 Table 与您的 Ajax 请求一起工作,您必须先调用请求,然后再使用数据 table
例子
请注意,在调用 ajax 之前不必先调用 DataTable,table 需要将数据输入 table 正文
<tbody class="load-transactions"> and append using jquery for the rest
试试这个。稍后谢谢我。
我正在努力实现 https://almsaeedstudio.com/themes/AdminLTE/pages/tables/data.html - "Data Table With Full Features"
当我静态添加 tbody 时,分页和排序工作正常,但是当我使用 jquery 添加 tbody 时,如下所示,添加了行但分页和排序失败。
HTML
<table id="tblItems">
<thead>
<tr>
<th>code</th>
<th>Name</th>
<th>Description</th>
<th>Image</th>
<th>Parent</th>
<th>Location</th>
<th>Category</th>
</tr>
</thead>
</table>
jquery
$(document).ready(function() {
$('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
$('#tblItems').append('<tbody><tr><td>asdsa34id</td><td> asdsa34id </td><td>asdsa34id </td><td> asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td></tr></tbody>');
});
https://jsfiddle.net/techjerk2013/vwpsxhaL/
更新代码
更新后的代码不会填充 table 尽管有来自响应的数据。虽然我将 deferRender 设置为 true,但数据table 仍然是空的。
$(document).ready(function() {
PopulateItemsTable();
BindTable();
});
function BindTable() {
$("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
function PopulateItemsTable() {
var txt = "";
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
if (jsonObject) {
var len = jsonObject.length;
if (len > 0) {
for (var i = 0; i < len; i++) {
if (jsonObject[i].Id) {
Id = jsonObject[i].Id;
}
else {
Id = '';
}
if (jsonObject[i].Name) {
Name = jsonObject[i].Name;
}
else {
Name = '';
}
if (jsonObject[i].Description) {
Description = jsonObject[i].Description;
}
else {
Description = '';
}
if (jsonObject[i].Image) {
Image = jsonObject[i].Image;
}
else {
Image = '';
}
if (jsonObject[i].Parent) {
Parent = jsonObject[i].Parent;
}
else {
Parent = '';
}
if (jsonObject[i].Location) {
Location = jsonObject[i].Location;
}
else {
Location = '';
}
Category = '';
txt += "<tr><td>" + Id + "</td><td>" + Name + "</td><td>" + Description + "</td><td>" + Image + "</td><td>" + Parent + "</td><td>" + Location + "</td><td>" + Category + "</td></tr>";
$('#tblItems').append('<tbody>' + txt + '</tbody>');
}
}
else {
$("#tblItems").append("No records found");
}
}
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
回答
在下面回答的人的帮助下,下面的代码按预期工作。
<script type="text/javascript">
var myTable;
$(document).ready(function () {
BindItemTable();
PopulateItemsTable();
});
function BindItemTable() {
myTable = $("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
function PopulateItemsTable() {
$.ajax({
type: "POST",
url: "ItemManagement.aspx/SearchIdList",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function (item) {
var result = [];
result.push(item.Id);
result.push(item.Name);
result.push(item.Description);
result.push(item.Image);
result.push(item.Parent);
result.push(item.Location);
result.push("");
return result;
});
myTable.rows.add(result);
myTable.draw();
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
</script>
不要直接将行添加到 table 标记,而是将其添加到 DataTable 实例,然后使用 .draw()
方法。无论如何,添加到 DataTable 实例都会在内部将其添加为 tbody
。像这样的事情应该做
var mytable = $('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
mytable.row.add(['asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id', 'asdsa34id']);
mytable.draw();
这是一个演示 https://jsfiddle.net/dhirajbodicherla/vwpsxhaL/1/
还阅读 how to add rows to DataTable 他们的文档以供进一步参考
更新
您可以使用 rows.add()
(复数)并执行类似这样的操作
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function(item){
var result = [];
result.push(item.Id);
// .... add all the values required
return result;
});
myTable.rows.add(result); // add to DataTable instance
myTable.draw(); // always redraw
var myTable;
$(document).ready(function() {
myTable = $("#tblItems").DataTable({
"deferRender": true,
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
PopulateItemsTable();
});
function PopulateItemsTable() {
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d);
var result = jsonObject.map(function(item){
var result = [];
result.push(item.Id);
// .... add all the values required
return result;
});
myTable.rows.add(result); // add to DataTable instance
myTable.draw(); // always redraw
}
});
}
如果您正在修改 table html 使用 jQuery 而不是插件提供的 api,那么您必须再次调用插件,以便它会根据修改后的内容重新实例化html.
$(document).ready(function() {
$('#tblItems').append('<tbody><tr><td>asdsa34id</td><td> asdsa34id </td><td>asdsa34id </td><td> asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td><td>asdsa34id</td></tr></tbody>');
$('#tblItems').DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
});
演示 https://jsfiddle.net/8hyr08xb/
根据编辑后的问题更新
试试这个
function PopulateItemsTable() {
var txt = "";
$.ajax({
type: "POST",
url: "Item.aspx/Search",
contentType: "application/json; charset=utf-8",
dataType: "json",
success: function (response) {
var jsonObject = JSON.parse(response.d), html = [];
if (jsonObject) {
var len = jsonObject.length;
if (len > 0) {
for (var i = 0; i < len; i++) {
if (jsonObject[i].Id) {
Id = jsonObject[i].Id;
}
else {
Id = '';
}
if (jsonObject[i].Name) {
Name = jsonObject[i].Name;
}
else {
Name = '';
}
if (jsonObject[i].Description) {
Description = jsonObject[i].Description;
}
else {
Description = '';
}
if (jsonObject[i].Image) {
Image = jsonObject[i].Image;
}
else {
Image = '';
}
if (jsonObject[i].Parent) {
Parent = jsonObject[i].Parent;
}
else {
Parent = '';
}
if (jsonObject[i].Location) {
Location = jsonObject[i].Location;
}
else {
Location = '';
}
Category = '';
html.push("<tr><td>" + Id + "</td><td>" + Name + "</td><td>" + Description + "</td><td>" + Image + "</td><td>" + Parent + "</td><td>" + Location + "</td><td>" + Category + "</td></tr>");
}
$('#tblItems')
.append('<tbody>' + html.join('') + '</tbody>')
.DataTable({
"paging": true,
"lengthChange": false,
"searching": false,
"ordering": true,
"info": true,
"autoWidth": false,
"sDom": 'lfrtip'
});
}
else {
$("#tblItems").append("No records found");
}
}
},
failure: function () {
$("#tblItems").append(" Error when fetching data please contact administrator");
}
});
}
相信我,我做了以上所有事情 none 无需编写长行代码就可以按预期工作
对我来说,解决方案非常简单快捷。 现在要让数据 Table 与您的 Ajax 请求一起工作,您必须先调用请求,然后再使用数据 table 例子
请注意,在调用 ajax 之前不必先调用 DataTable,table 需要将数据输入 table 正文
<tbody class="load-transactions"> and append using jquery for the rest
试试这个。稍后谢谢我。