非常直截了当。在javascript中,我需要检查字符串是否包含数组中持有的任何子字符串。


当前回答

如果您正在处理一长串子字符串,这些子字符串由空格或任何其他常见字符分隔的完整“单词”组成,那么您可以在搜索中使用一些聪明的方法。

首先把你的字符串分成X组,然后X+1,然后X+2,…X和Y应分别为子字符串中单词最少和最多的单词数。例如,如果X是1,Y是4,“Alpha Beta Gamma Delta”变成:

"阿尔法" "贝塔" "伽马" "德尔塔"

"阿尔法·贝塔" "贝塔·伽马" "伽马·德尔塔"

"Alpha Beta Gamma" "Beta Gamma Delta"

"Alpha Beta Gamma Delta"

如果X是2 Y是3,那么你可以省略第一行和最后一行。

现在,如果您将它插入到Set(或Map)中,就可以快速搜索这个列表,比字符串比较快得多。

缺点是你不能搜索像“ta Gamm”这样的子字符串。当然,你可以通过按字符而不是按单词进行分割来实现这一点,但这样你通常需要构建一个庞大的Set,并且这样做所花费的时间/内存超过了好处。

其他回答

借鉴T.J. Crowder的解决方案,我创建了一个原型来处理这个问题:

Array.prototype.check = function (s) {
  return this.some((v) => {
    return s.indexOf(v) >= 0;
  });
};
function containsAny(str, substrings) {
    for (var i = 0; i != substrings.length; i++) {
       var substring = substrings[i];
       if (str.indexOf(substring) != - 1) {
         return substring;
       }
    }
    return null; 
}

var result = containsAny("defg", ["ab", "cd", "ef"]);
console.log("String was found in substring " + result);
var yourstring = 'tasty food'; // the string to check against


var substrings = ['foo','bar'],
    length = substrings.length;
while(length--) {
   if (yourstring.indexOf(substrings[length])!=-1) {
       // one of the substrings is in yourstring
   }
}

我并不是建议你去扩展/修改String的原型,但这是我所做的:

String.prototype.includes ()

String.prototype.includes = function (includes) { console.warn("String.prototype.includes() has been modified."); return function (searchString, position) { if (searchString instanceof Array) { for (var i = 0; i < searchString.length; i++) { if (includes.call(this, searchString[i], position)) { return true; } } return false; } else { return includes.call(this, searchString, position); } } }(String.prototype.includes); console.log('"Hello, World!".includes("foo");', "Hello, World!".includes("foo") ); // false console.log('"Hello, World!".includes(",");', "Hello, World!".includes(",") ); // true console.log('"Hello, World!".includes(["foo", ","])', "Hello, World!".includes(["foo", ","]) ); // true console.log('"Hello, World!".includes(["foo", ","], 6)', "Hello, World!".includes(["foo", ","], 6) ); // false

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = 0, len = arr.length; i < len; ++i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}

编辑: 如果测试的顺序不重要,你可以使用这个(只有一个循环变量):

var str = "texttexttext";
var arr = ["asd", "ghj", "xtte"];
for (var i = arr.length - 1; i >= 0; --i) {
    if (str.indexOf(arr[i]) != -1) {
        // str contains arr[i]
    }
}