1
我有一個方法是採取簽名中的對象的參數。我想傳遞該對象而不是許多參數,另一方面,我不想更改現有方法的簽名,因爲它已在多個位置使用。所以基本上我想要兩種方法。但是當我試圖編寫代碼時,它給了我錯誤Duplicate function implementation
。如何在TypeScript中重載方法?
getSearchData(fetchData: FetchData,languageCode: string, sorting: string, maxResultCount: number, skipCount: number): Observable<PagedResultDtoOfFetchData> {
getSearchData(dataLevel: number, codeType: number, dataCode: string, descLong: string, languageCode: string, dataParent: string, sorting: string, maxResultCount: number, skipCount: number): Observable<PagedResultDtoOfFetchData> {
FYI dataLevel, codeType, dataCode, descLong, dataParent
是fetchData
性質。
文檔 - https://www.typescriptlang.org/docs/handbook/functions.html#overloads –
[打字稿函數重載](的可能的複製https://stackoverflow.com/questions/13212625/typescript-function -overloading) – marvinhagemeister
@gsamaras你爲什麼這麼貪婪? –