如何降低 JavaScript 中 if else 语句的复杂性？

Question

我有以下场景，它有多个 if else 条件。

以下代码的圈复杂度显示为 7。

有没有更好的方法来使用 Javascript 编写下面的代码 sn-p 来降低代码的复杂性？

function setTime() {

var currentTime = "3/4/2020, 2:53:42 PM"
var selectedTime = "3/5/2020, 2:53:42 PM"

if( Date.parse(currentTime) < Date.parse(selectedTime)) {
    callThisMethod('Current time less than selected time');
} else if (Date.parse(currentTime) > Date.parse(selectedTime)) {
    callThisMethod('Current time Greater than selected time');
} else {
    callThisMethod('Current time is equal to selected time');
}
}

function callThisMethod(message) {
 console.log(message);
}

setTime();

Answer 1

可能的选项之一：

const currentTime = new Date("3/4/2020, 2:53:42 PM"),
      selectedTime = new Date("3/5/2020, 2:53:42 PM")

callThisMethod(`Current time is ${currentTime < selectedTime ? 'less than' : currentTime > selectedTime ? 'greater than' : 'equal to'} selected time`)

Answer 2

您可以先获取 ISO 日期，然后比较此日期。

function setTime() {
  var currentTime = "3/4/2020, 2:53:42 PM",
      selectedTime = "3/5/2020, 2:53:42 PM",
      current = new Date(currentTime).toISOString(),
      selected =  new Date(selectedTime).toISOString();

  if (current < selected) {
    callThisMethod('Current time less than selected time');
  } else if (current > selected) {
    callThisMethod('Current time Greater than selected time');
  } else {
    callThisMethod('Current time is equal to selected time');
  }
}

function callThisMethod(message) {
  console.log(message);
}

setTime();

Answer 3

function setTime() {

    var currentTime = "3/4/2020, 2:53:42 PM";
    var selectedTime = "3/5/2020, 2:53:42 PM";
    let a = new Date(currentTime).getTime();
    let b = new Date(selectedTime).getTime();
    let str = ['less than','is equal','greater than'];
    let n = (a-b)/Math.abs(a-b) || 0;

    callThisMethod(`Current time ${str[n]} selected time`);
}

function callThisMethod(message) {
 console.log(message);
}

setTime();

根据https://jshint.com/，这个函数的圈复杂度数是2。

Answer 4

也许你应该写“减少代码”或“制作更好的代码”。这个 if 或 else if 语句并不复杂。

其实，你为什么不 console.log 来自 if/else if 的消息，你避免调用另一个函数？

function setTime() {
      var currentTime = "3/4/2020, 2:53:42 PM",
          selectedTime = "3/5/2020, 2:53:42 PM",
          current = new Date(currentTime).toISOString(),
          selected =  new Date(selectedTime).toISOString();

      if (current < selected) {
        console.log('Current time less than selected time');
      } else if (current > selected) {
        console.log('Current time Greater than selected time')
      } else {
        console.log('Current time is equal to selected time');
      }
    }

    setTime();

Answer 5

这种方法可能有几种可能的答案。 您可以做的一件事是使条件免于评估。因此，不是在条件表达式中解析日期，而是在比较之前解析它。之后，您可以将最常见的条件（您认为可能经常出现的情况）作为第一个条件，并使用二进制搜索方法打破其他条件。考虑以下代码 sn-p：

function setTime() {
    var currentTime = Date.parse("3/4/2020, 2:53:42 PM")
    var selectedTime = Date.parse("3/5/2020, 2:53:42 PM")
    if(currentTime < selectedTime) {
       callThisMethod('Current time less than selected time');
    } else {
       if (currentTime > selectedTime) {
         callThisMethod('Current time Greater than selected time');
       } else {
          callThisMethod('Current time is equal to selected time');
         }
      }
}

function callThisMethod(message) {
 console.log(message);
}

setTime();

注意条件句是如何使用拆分技术分解的。 或者你可以使用 switch case 而不是条件。与 if-else 条件相比，已发现 Switch case 在后续条件上的增量成本更少。 考虑以下 sn-p：

function setTime() {
        var currentTime = Date.parse("3/4/2020, 2:53:42 PM")
        var selectedTime = Date.parse("3/5/2020, 2:53:42 PM")
        switch(true){
            case (currentTime > selectedTime):
                callThisMethod('Current time Greater than selected time');
                break;
            case (currentTime < selectedTime):
                callThisMethod('Current time less than selected time')
                break;
            default:
                 callThisMethod('Current time is equal to selected time')    
}
    }

    function callThisMethod(message) {
     console.log(message);
    }

    setTime();