我的页面中有一个剑道 ui 网格,其中包含一些列。现在我想添加一列来显示我的行号。我该怎么做?谢谢。
问问题
26167 次
8 回答
26
初始化一个变量并在列中显示为template: "#= ++record #"
这是代码:
var record = 0;
$("#grid").kendoGrid({
dataSource: {
data: [{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" },{ foo: "foo" }, { foo: "foo" }, { foo : "foo1" }],
pageSize: 10
},
sortable: true,
columns: [ {
title: " ",
template: "#= ++record #",
width: 30
}, { field: "foo" }],
pageable: true,
dataBinding: function() {
record = (this.dataSource.page() -1) * this.dataSource.pageSize();
}
});
于 2013-06-29T08:53:51.473 回答
7
对于 Razor,您还需要实际显示数字:(没有足够的点数来评论)
网格上方:
@{int counter = 1;}
内列定义:
columns.Template(@<text>
<span>@counter @{ counter++; }</span>
</text>).Title("#");
于 2013-10-10T20:45:59.640 回答
7
无需定义任何变量,您可以从数据绑定事件中获得帮助:
$("#grid").kendoGrid({
sortable: true,
dataSource: [{
name: "Jane Doe",
age: 30
}, {
name: "John Doe",
age: 33
}],
columns: [{
field: "name"
}, {
field: "age"
}, {
field: "rowNumber",
title: "Row number",
template: "<span class='row-number'></span>"
}],
dataBound: function () {
var rows = this.items();
$(rows).each(function () {
var index = $(this).index() + 1
+ ($("#grid").data("kendoGrid").dataSource.pageSize() * ($("#grid").data("kendoGrid").dataSource.page() - 1));;
var rowLabel = $(this).find(".row-number");
$(rowLabel).html(index);
});
}
});
于 2015-11-12T14:01:39.540 回答
4
YD1m 的模板对我不起作用,它将变量视为string
. 所以我不得不像这样实现它:
columns.Template(@<text>@((long)counter++)</text>)
于 2013-07-16T14:20:01.197 回答
3
对于 asp.net mvc 包装器,您应该使用如下内容:
@{
var counter = 1;
}
@( Html.Kendo().Grid<Types>()
.Name("grid")
.Columns(columns =>
{
// Define a template column with row counter
columns.Template(@<text>@counter++</text>);
// Define a columns from your data set and set a column setting
columns.Bound(type => type.id);
columns.Bound(type=> type.name).Title("Name");
// add more columns here
})
)
于 2013-06-29T08:55:32.063 回答
0
如果需要,可编辑网格中的行号
$(document).ready(function(){
$("#grid").kendoGrid({
dataSource: {
data: null,
schema: {
model: {
id: "register_No",
fields: {
register_No: {editable: true},
myEditableField: {editable: true},
}
}
}
},
edit:function(e){
var fields= $('input[name="register_No"]');
fields.attr('disabled', true);
fields.removeClass('k-input k-textbox');
fields.addClass('none');
//create this class and set border and background none
$('input[name="myEditableField"]').focus();
},
save:function(e){
var total=e.sender.dataSource.data().length;
if(e.model.register_No==""){
e.model.register_No=total;
}
},
remove:function(e){
var grid = $("#grid").data("kendoGrid");
var todos=grid.dataSource.data();
var id_actual=e.model.register_No;
var count=1;
for(var i=0;i<todos.length;i++){
if(todos[i].register_No!==id_actual){
var data = grid.dataSource.at(i);
data.set("register_No", count);
count++;
}
}
}
, height: 280,
toolbar: ["create"],
scrollable: true,
editable: {mode:"inline", createAt: "bottom", confirmation: true
}
,
columns: [
{field: "register_No",title: "No", width: 30},
{field: "myEditableField", title: "Any Field", width: 120},
{field: "options", command: ["edit", "destroy"], width: 200}
]
});
});
<div id="grid"></div>
于 2016-06-09T17:55:47.350 回答
0
声明一个用于行计数的变量:
var rowNumber = 0;
在模板中使用 ++ 运算符来为每次迭代增加它:
#= ++rowNumber #
这也适用于 Kendo UI ListView。
见官方文档:
http://docs.telerik.com/kendo-ui/controls/data-management/grid/how-to/Templates/add-row-numbers
于 2017-04-11T20:51:30.747 回答
0
基于 Ehsan Mirsaeedi 的出色回答,我提出了这个版本,它分配从 0 开始的索引而不是从 1 开始的行号,如果网格被分组,则跳过组标题,并处理网格未分页的情况。
我需要这些索引来为每一列添加一个带有隐藏输入的模板,这样我就可以将网格的值与整个表单一起提交。
我认为它足够相关,值得添加到线程中......
代码:
var theGrid = $("#grid").data("kendoGrid");
var rows = this.items().filter(function (index, item) {
return $(item).hasClass("k-grouping-row") == false;
});
$(rows).each(function () {
var index = $(this).index();
//prevent group header rows from incrementing index
if (theGrid.dataSource.options.group != null && theGrid.dataSource.options.group.length > 0)
index -= $(this).prevAll("#grid tr.k-grouping-row").length;
//adjust indices for current page
if (theGrid.options.pageable == true)
index += theGrid.dataSource.pageSize() * (theGrid.dataSource.page() - 1);
//add the value to the grid's data object
theGrid.dataItem(this).rowNumber = index;
//and update the grid's HTML
var rowLabel = $(this).find(".row-number");
$(rowLabel).html(index);
});
于 2016-04-19T08:06:00.877 回答