如何在过滤器方法中返回与&&在回调函数中加入的布尔值？

Question

我正在寻找一种生成布尔值的优雅方法，该布尔值最终将在过滤器方法的回调函数内使用&&运算符加入。我试图遍历过滤条件，但是找不到将每个迭代结果合并为以下格式的方法：return Boolean && Boolean && Boolean && Boolean && Boolean因为+ = &&布尔无效。这是我所拥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})我需要可扩展且更优雅的解决方案，因此，如果我增强了过滤后的数组，则无需添加&&。

Answer 1

您可以通过应用Array.every()并String.includes()像这样来做到这一点：

var searchdata = this.preSearch.filter(row => {

    // this only returns true if our condition works for

    // index = 0, 1, 2, 3, 4

    return [0, 1, 2, 3, 4].every(index => {

        const rowContent = row[index].toLowerCase();

        const filterContent = filters[index].toLowerCase();

        // String.includes() is nicer than String.indexOf() here because

        // you don't need the ugly -1

        return rowContent.includes(filterContent);

    });

});