javascript - 如何实现循环额外条件

Question

我试图实现一些非常简单的事情，但找不到正确的解决方案:如果在范围内找到，如何替换字符串中的符号，如果不在范围内，则如何保持它们“原样”。

    var BASE = "ABCDEFGHIJ"; 
    var CODE = "0123456789";
    var WORD = "DEKF" // K is out of range
    var CONS = []; // result expected: 34K5

    for (b=0; b<BASE.length; b++){

        for (w=0; w<WORD.length; w++){

            // if a sign of WORD is in BASE we change it with CODE equivalent: A->0
            if(WORD[w]==BASE[b]) {
                CONS.push(CODE[b]); 
            }

            // if not we keep it in place 
            else {
                // make sure the sign is out of BASE range
                if(BASE.search(WORD[w]) == -1 ) { 
                    CONS.push(WORD[w]); 
                    break;
                }
            }           
        }
    }

    console.log(CONS.join(''))

控制台

without the else condition = 345 ( no K )
else cond without break = KKK3K4KK5KKKK
else cond with break = KKK3K4KKKKKK
else without if->match and without break = DEKFDEKFDEKF3EKFD4KFDEK5DEKFDEKFDEKFDEKF
else without if->match and with break = DDD3EDDDDDD
with continue instead of break =  KKK3K4KK5KKKK

然后我尝试使用标签来打破循环:

    for (b=0; b<BASE.length; b++){

        sign:
        for (w=0; w<WORD.length; w++){

            // if a sign of WORD is in BASE we change it with CODE equivalent: A -> 0
            if(WORD[w]==BASE[b]) {
                CONS.push(CODE[b]); 
            }

            // if not we keep it in place: 
            else {
                if(BASE.search(WORD[w]) == -1 ) { // make sure the sign is out of BASE range
                    CONS.push(WORD[w]); 
                    break sign;
                }
            }           
        }
    }

控制台

with sign label and break sign = KKK3K4KKKKKK
with sign label before the first loop and break sign; = K

Answer 1

提供替代方案。特别是如果列表变得更长并且为了可维护性，您可以使用 Map 而不是 2 个松散的字符串。(在下面的示例中， map 是从字符串创建的)

之后，您可以简单地将所有字符替换为 map 的条目或字符本身:

const BASE = "ABCDEFGHIJ", CODE = "0123456789", WORD = "DEKF",
	codes = new Map([...BASE].map((b,i)=> [b,CODE[i]]));
    
let CONS = [...WORD].map(s=> codes.get(s) || s);

console.log(CONS.join(''));