Question

过滤URL列表的简洁，面向功能的方法是什么，其中每个项目必须传递一系列测试？如果URL匹配任何测试，则应将其过滤掉。

我目前有：

var _ = require("underscore");

const anchors = [
    {href:"https://example.org/contact"},
    {href:"https://example.org/faq"},
    {href:"https://example.org/contact"},
    {href:"https://example.org/uploads/image-1024x1018.jpg"},
    {href:"https://example.org/wp-json/oembed/1.0/embed?url=example"},
    {href:"https://example.org/author/pm"},
    {href:"https://example.org/wp/wp-login.php?action=lostpassword"},
    {href:"https://example.org/wp/wp-login.php"},
    {href:"https://example.org/feed"},
];

const tests = [
    /\/wp\//,
    /\/wp-json\//,
    /\.jpg$/,
    /\.png$/,
    /\.gif$/,
]

function testAll(testString){
    let pass = true;
    _.each(tests, t => {
        if(t.test(testString)) pass = false;
    });
    return pass;
}

console.log(anchors.map(anchor => {
    return anchor.href;
}).filter(anchor => {
    return testAll(anchor);
}));

但我怀疑testAll可以更简洁的方式完成。

Answer 1

我正在寻找的解决方案实际上是some而不是every，因为如果匹配任何的测试，我实际上需要拒绝该网址：

console.log(anchors.map(anchor => {
    return anchor.href;
}).filter(anchor => {
    // return testAll(anchor);
    return !_.some(tests, t => {
        return t.test(anchor);
    })
}));

Answer 2

您可以使用Array#every()

function testAll(testString){
    return tests.every(reg => reg.test(testString));
}

Answer 3

您可以使用Array#some并将检查的否定结果用于过滤。

var anchors = [{ href:"https://example.org/contact" }, { href:"https://example.org/faq" }, { href:"https://example.org/contact" }, { href:"https://example.org/uploads/image-1024x1018.jpg" }, { href:"https://example.org/wp-json/oembed/1.0/embed?url=example" }, { href:"https://example.org/author/pm" }, { href:"https://example.org/wp/wp-login.php?action=lostpassword" }, { href:"https://example.org/wp/wp-login.php" }, { href:"https://example.org/feed" }],
    tests = [/\/wp\//, /\/wp-json\//, /\.jpg$/, /\.png$/, /\.gif$/],
    result = anchors.filter(({ href }) => !tests.some(t => t.test(href)));

console.log(result);

按测试列表过滤

3 个答案: