[英]How to reduce the complexity of this if else statements in Javascript?
我有以下场景,其中有多个 if else 条件。
以下代码的圈复杂度显示为 7。
是否有更好的方法来使用 Javascript 编写以下代码片段以降低代码的复杂性?
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();
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`)
您可以先获取 ISO 日期,然后比较这些日期。
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();
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/ ,此函数的圈复杂度数为 2。
也许您应该编写“以减少代码”或“编写更好的代码”。 这个 if 或 else if 语句并不复杂。
实际上,为什么不从 if/else if 中 console.log 消息,并且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();
这种方法可能有几种可能的答案。 您可以做的一件事是保持条件不受评估。 因此,与其在条件表达式中解析日期,不如在比较之前解析它。 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,并使用二分搜索方法打破其他条件。 考虑以下代码片段:
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();
请注意如何使用拆分技术将条件分解为多个部分。 或者您可以使用 switch case 而不是条件。 与 if-else 条件相比,已发现 switch case 在后续条件上花费的增量成本更少。 考虑以下片段:
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();
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