我正在尋找一種生成布爾值的優(yōu)雅方法，該布爾值最終將在過濾器方法的回調(diào)函數(shù)內(nèi)使用&&運算符加入。我試圖遍歷過濾條件，但是找不到將每個迭代結(jié)果合并為以下格式的方法：return Boolean && Boolean && Boolean && Boolean && Boolean因為+ = &&布爾無效。這是我所擁有的并且正在起作用：//data I am filteringthis.preSearch = [  ["The Lord of the Rings", "J. R. R. Tolkien", "English", "1955", "150 milionów"],  ["Le Petit Prince (The Little Prince)", "Antoine de Saint-Exupéry", "French", "1943", "140 milionów"],  ["Harry Potter and the Philosopher's Stone", "J. K. Rowling", "English",  "1997", "120 milionów"],  ["The Hobbit", "J. R. R. Tolkien", "English", "1937", "100 milionów"],  ["And Then There Were None", "Agatha Christie",   "English", "1939",  "100 milionów"],  ["Dream of the Red Chamber",  "Cao Xueqin",   "Chinese", "1791", "100 milionów"]]//filters, that are set dynamically but let's pretend they are equal tovar filters = ["", "", "english", "19", "1"]var searchdata = this.preSearch.filter(row => {          return     row[0].toLowerCase().indexOf(filters[0].toLowerCase()) > -1     && row[1].toLowerCase().indexOf(filters[1].toLowerCase()) > -1     && row[2].toLowerCase().indexOf(filters[2].toLowerCase()) > -1     && row[3].toLowerCase().indexOf(filters[3].toLowerCase()) > -1     && row[4].toLowerCase().indexOf(filters[4].toLowerCase()) > -1})我需要可擴展且更優(yōu)雅的解決方案，因此，如果我增強了過濾后的數(shù)組，則無需添加&&。