14

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。

4

8 回答 8

26

初始化一个变量并在列中显示为template: "#= ++record #"

工作演示

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);
});

结果:
从零开始的行号跳过组标题

于 2016-04-19T08:06:00.877 回答