antd里面的Table组件，dataSource如果涉及children无法筛选里面的数据？-灵析社区

排序的话，实现起来挺简单，直接在sorter方法里面判断就行，发现筛选onFilter方法里面判断行不通，办法是有，我感觉很复杂，需要跟Table里面的onchange配合使用，得写一堆逻辑，还容易出问题，请问有其它更简单方法实现吗？比如直接在onFilter方法里面实现就行的  onFilter方法是一个布尔值,如果在children里面判断发现行不通 比如： onFilter: (value, record) => { if(record.children){ //xxx } } demo const { createRoot } = ReactDOM; const { Space, Switch, Table } = antd; const { useState } = React;; const columns = [ { title: 'Name', dataIndex: 'name', key: 'name', filters: [ { text: 'test01', value: 'test01', },{ text: 'test02', value: 'test02', }, { text: 'test10', value: 'test10', },{ text: 'test21', value: 'test21', }, ], onFilter: (value: string, record) => record.name === value }, { title: 'Age', dataIndex: 'age', key: 'age', width: '12%', }, { title: 'Address', dataIndex: 'address', width: '30%', key: 'address', }, ]; const data = [ { key: 1, name: '测试A', age: 60, address: 'New York No. 1 Lake Park', children: [ { key: 11, name: 'test02', age: 42, address: 'New York No. 2 Lake Park', },{ key: 12, name: 'test01', age: 22, address: 'New York No. 2 Lake Park', }, ], }, { key: 2, name: '测试A_1', age: 29, address: 'Sidney No. 1 Lake Park', }, { key: 3, name: '测试B', age: 32, address: 'Sidney No. 1 Lake Park', children: [ { key: 13, name: 'test21', age: 25, address: 'New York No. 2 Lake Park', },{ key: 14, name: 'test01', age: 19, address: 'New York No. 2 Lake Park', },{ key: 15, name: 'test10', age: 19, address: 'New York No. 2 Lake Park', }, ], },{ key: 4, name: '测试B_1', age: 29, address: 'Sidney No. 1 Lake Park', }, ]; const App = () => { return ( ); }; const ComponentDemo = App; createRoot(mountNode).render();