测试动态创建的字符串

Question

我现在拥有的是一个函数，返回使用随机数生成器放在一起的2个字符串的数组。这些值将帮助组成一个字符串，如“这是”+ v1 +“，这是”+ v2。 v1和v2是动态创建的字符串。我需要根据字符串的值进行操作。我的问题是我不知道如何对v1和v2所代表的所有不同方式进行有效比较。我想我必须做很多if语句，就像我将在代码中显示的方式一样。我知道有一个更好的方法来进行比较。它是什么？

//the unit function makes sure that a random letter is returned but also makes sure that 
    the first letter returned comes before the second letter in the values array 
function unit(){ 
     var values = ["a", "b", "c", "d", "e" , "f"]; 
     var value1 = Math.floor(Math.random() * (values.length - 1)) + 1; 
     var value2 = Math.floor(Math.random() * value1); 
     if(value1 !== value2){ 
      return [values[value2], values[value1]]; 
     } 

    } 
    //pseusodo-code 
     // i have to do a different operation dependeng on the values of v1 and v2 
     //bad code 
    function conversion(v1, v2, num){ 
     if(v1 == "a" && v2 == "b"){ 
      return 16 * num 
     } 
     if(v1 == "a" && v2 == "c"){ 
      return 32 * num 
     } 
     if(v1 == "a" && v2 == "d"){ 
      return 64 * num 
     } 
     if(v1 == "a" && v2 == "e"){ 
      return 256 * num 
     } 
     if(v1 == "b" && v2 == "c"){ 
      return 18 * num 
     } 
     if(v1 == "b" && v2 == "d"){ 
      return 20 * num 
     } 
     if(v1 == "b" && v2 == "e"){ 
      return 64 * num 
     } 
     if(v1 == "b" && v2 == "f"){ 
      return 256 * num 
     } 
     etc 
    } 
    function string(){ 
     u = unit() 
     v1 = u[0]; 
     v2 = u[1] 
     return "this is " + v1 + " and this " + v2 
    } 
    console.log(string()) 

    // for every 2 a theres 1 b's 
    //for every 16 a there 1 c 
    //for every 32 a there is 1 d 
    //for every 64 a there is 1 e 
    conversion(v1,v2, 2.5) 
    console.log(v1, " ", v2) 

Answer 1

难道你不能创建一个查找对象，其中包含所有的乘法因子？ 想到这好像它是一个二维表：

var factors = { 
    a: { b: 16, c: 32, d: 64, e: 256 }, 
    b: { c: 18, d: 20, e: 64, f: 256 } 
}; 

然后转换函数变为：

function conversion (v1, v2, num) { 
    if (! factors.hasOwnProperty(v1)) { 
    throw "no conversion defined on first level for value " + v1; 
    } 
    if (! factors[v1].hasOwnProperty(v2)) { 
    throw "no conversion defined on second level for value " + v1 + ", " + v2; 
    } 
    return num * factors[v1][v2]; 
} 

注意，在factors对象而我的手编码上述值，其值也可以以编程方式设置。

Answer 2

你就可以用这两个值的开关语句作为字符串，像这样：

function conversion(v1,v2,num){ 
    var switchVar = v1 + "," + v2; // or use another separator than comma, depending of your case 
    int m = 0; 
    switch(switchVar){ 
    case "a,b": 
     m = 16; 
     break; 
    case "a,c": 
    ... 
    default: 
     break; 
    } 
    return i*num; 
}