107 lines
3.1 KiB
TypeScript
107 lines
3.1 KiB
TypeScript
import { Staff } from "@prisma/client";
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import { DataNode } from "./type";
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export function findNodeByKey(
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nodes: DataNode[],
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targetKey: string
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): DataNode | null {
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let result: DataNode | null = null;
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for (const node of nodes) {
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if (node.key === targetKey) {
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return node;
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}
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if (node.children && node.children.length > 0) {
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result = findNodeByKey(node.children, targetKey);
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if (result) {
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return result;
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}
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}
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}
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return result;
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}
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export function findStaffById(
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nodes: DataNode[],
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staffId: string
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): Staff | null {
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for (const node of nodes) {
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// 在当前节点的staffs数组中查找
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const foundStaff = node?.data?.staffs.find(
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(staff: Staff) => staff.id === staffId
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);
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if (foundStaff) {
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return foundStaff;
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}
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// 如果当前节点的staffs数组中没有找到，则递归在子节点中查找
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if (node.children) {
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const foundInChildren = findStaffById(node.children, staffId);
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if (foundInChildren) {
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return foundInChildren;
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}
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}
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}
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// 如果在所有节点及其子节点中都没有找到，返回null
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return null;
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}
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interface MappingConfig {
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titleField?: string;
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keyField?: string;
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valueField?: string;
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hasChildrenField?: string; // Optional, in case the structure has nested items
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childrenField?: string;
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}
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export function mapToDataNodes(
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inputArray: any[],
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config: MappingConfig = {}
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): DataNode[] {
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const {
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titleField = "title",
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keyField = "key",
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valueField = "value",
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hasChildrenField = "hasChildren",
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childrenField = "children"
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} = config;
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return inputArray.map((item) => {
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const hasChildren = item[hasChildrenField] || false;
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const children = item[childrenField]
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return {
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title: item[titleField] || "",
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key: item[keyField] || "",
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value: item[valueField] || null,
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data: item,
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children: children
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? mapToDataNodes(children, { titleField, keyField, valueField, hasChildrenField, childrenField })
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: undefined,
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hasChildren
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};
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});
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}
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/**
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* 合并两个数组并去重。
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*
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* 该函数将两个输入数组的元素合并为一个数组，
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* 并确保结果数组中没有重复的元素。元素的顺序根据它们首次出现的顺序保留。
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*
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* @template T - 输入数组中元素的类型。
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* @param {T[]} array1 - 要合并的第一个数组。
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* @param {T[]} array2 - 要合并的第二个数组。
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* @returns {T[]} 包含来自两个输入数组的唯一元素的新数组。
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*
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* @example
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* const array1 = [1, 2, 3, 4];
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* const array2 = [3, 4, 5, 6];
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* const result = mergeAndDeduplicate(array1, array2);
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* console.log(result); // 输出: [1, 2, 3, 4, 5, 6]
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*/
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export function mergeAndDeduplicate<T = string>(array1: T[], array2: T[]): T[] {
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const set = new Set([...array1, ...array2]);
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return Array.from(set);
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}
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