Question

我有以下简单的代码：

var S = require('string');

function matchBlacklist(inputString) {
    var blacklist = ["facebook", "wikipedia", "search.ch", "local.ch"];
    var found = false;
    for (var i = 0; i < blacklist.length; i++) {

        if (S(inputString).contains(blacklist[i]) > -1) {
            found = true;
        }
    }
    return (found);
}

matchBlacklist("www.facebook.com/this_is_a_test"); // returns true
matchBlacklist("www.example.com/this_is_a_test"); // returns true

但它总是回归真实;因为它应该为第二种情况返回false

Answer 1

您应该测试if(string.includes(substring)){ ... }而不是if(string.includes(substring) > -1 ){ ... }

但这是一个更优雅的单行：

const blacklist = ["facebook", "wikipedia", "search.ch", "local.ch"];

const matchBlacklist = inputString => blacklist.some(word => inputString.includes(word))

console.log( matchBlacklist("www.facebook.com/this_is_a_test") ); // returns true
console.log( matchBlacklist("www.example.com/this_is_a_test") ); // returns false

Answer 2

使用数组find方法的更好方法，如果表达式求值true否则返回元素，则返回基于查找结果的布尔值

Find method MDN

Includes method MDN

function matchBlacklist(inputString) {
    var blacklist = ["facebook", "wikipedia", "search.ch", "local.ch"];
    
    return !!blacklist.find(b => inputString.includes(b))
}

console.log(matchBlacklist("www.facebook.com/this_is_a_test"));
console.log(matchBlacklist("www.example.com/this_is_a_test"));

Answer 3

您可以使用“indexOf”代替“contains”方法。 “contains”方法将返回布尔值，因此您无法将布尔值与数字进行比较。 “indexOf”将返回字符串的索引号，因此您可以在此处进行比较。

var S = require('string');

function matchBlacklist(inputString) {
  var blacklist = ["facebook", "wikipedia", "search.ch", "local.ch"];
  var found = false;
  for (var i = 0; i < blacklist.length; i++) {

    if (S(inputString).indexOf(blacklist[i]) > -1) {
        found = true;
    }
  }
  return (found);

}

matchBlacklist("www.facebook.com/this_is_a_test"); // returns true
matchBlacklist("www.example.com/this_is_a_test"); // returns truetes

使用nodeJS在URL中查找子字符串

3 个答案: