Question

我有一个带有工具栏模板的Telerik Kendo Grid，用于创建个性化过滤器。

我的工具栏中有2个下拉列表，值选择必须在＆＃34;和＆＃34;使用所有过滤器（包括网格的默认过滤器）。

当我想删除过滤器时，我使用此功能：

CODE

<script type="text/javascript">
 function removeFilter(filter, searchFor) {
    if (filter == null)
        return [];

    for (var x = 0; x < filter.length; x++) {

        if (filter[x].filters != null && filter[x].filters.length >= 0) {
            if (filter[x].filters.length == 0) {
                filter.splice(x, 1);
                return removeFilter(filter, searchFor);
            }
            filter[x].filters = removeFilter(filter[x].filters, searchFor);
        }
        else {
            if (filter[x].field == searchFor) {
                filter.splice(x, 1);
                return removeFilter(filter, searchFor);
            }
        }
    }

    return filter;
}
</script>

我的问题是我的函数removeFilter删除了gridview中的所有过滤器，而我只删除了特定的字段。

我在jsfiddle中重现Example。

问题

从网格过滤器中仅删除特定字段的正确方法是什么？

参考

GRID / Toolbar template

Answer 1

你的错误在myFilterChange function。您在更改时添加过滤器，但从不删除上一个过滤器。在此结果中，在第二次更改此组合框后，您的条件如idcat == 1 && idcat == 2。

您应该在为同一列fiddle添加另一个列之前清除过滤器：

if (statoFilterValue) {
    removeFilter(currFilters.filters, "idstate");
    currFilters.filters.push({ field: "idstate", operator: "eq", value: parseInt(statoFilterValue) });
} else {
    removeFilter(currFilters.filters, "idstate");
}

同样在removeFilter function中，您不应该使用递归与拼接相结合。足以确定splice

的结果

if (filter[x].filters.length == 0) {
   console.log('empty');
   filter = filter.splice(x, 1);
}

Answer 2

按文件名删除过滤器

    function removefilter(filterField)
{
    //grid id
    var gridData = $("#your  grid  id").data("kendoGrid");

    // get currently applied filters from the Grid.
    var currFilterObj = gridData.dataSource.filter();

    // get current set of filters, which is supposed to be array.
    // if the oject we obtained above is null/undefined, set this to an empty array
    var currentFilters = currFilterObj ? currFilterObj.filters : [];

    // iterate over current filters array. if a filter for "filterField" is already
    // remove  filter  by  filed name
    if (currentFilters && currentFilters.length > 0) {
        for (var i = 0; i < currentFilters.length; i++) {
            //  for(var j=0; j<)
            if (currentFilters[i].field == filterField) {
                currentFilters.splice(i, 1);
                break;
            }
        }
    }



    gridData.dataSource.filter({
        logic: "and",
        filters: currentFilters
    });
}

添加过滤器

    function applyFilter(filterField, filterValue) {

    // your  grid  id
    var gridData = $("#grid  id").data("kendoGrid");

    // get currently applied filters from the Grid.
    var currFilterObj = gridData.dataSource.filter();

    // get current set of filters, which is supposed to be array.
    // if the oject we obtained above is null/undefined, set this to an empty array
    var currentFilters = currFilterObj ? currFilterObj.filters : [];

    // iterate over current filters array. if a filter for "filterField" is already
    // defined, remove it from the array
    // once an entry is removed, we stop looking at the rest of the array.
    if (currentFilters && currentFilters.length > 0) {
        for (var i = 0; i < currentFilters.length; i++) {
            if (currentFilters[i].field == filterField) {
                currentFilters.splice(i, 1);
                break;
            }
        }
    }

    // if "filterValue" is "0", meaning "-- select --" option is selected, we don't 
    // do any further processing. That will be equivalent of removing the filter.
    // if a filterValue is selected, we add a new object to the currentFilters array.
    if (filterValue != "0") {
        currentFilters.push({
            field: filterField,
            operator: "eq",
            value: filterValue
        });
    }
    debugger;
    // finally, the currentFilters array is applied back to the Grid, using "and" logic.
    gridData.dataSource.filter({
        logic: "and",
        filters: currentFilters
    });

}

基于此enter link description here

使用工具栏模板删除Kendo Grid中的特定过滤器

2 个答案: