对我来说，这是最简单的方法:

switch (["afshin","saeed","larry"].includes(varName) ? 1 : 2) {
   case 1:
       alert('Hey');
       break;

   default:
       alert('Default case');
       break;
}

我可以看到这里有很多很好的答案，但如果我们需要检查10个以上的情况怎么办?以下是我自己的方法:

 function isAccessible(varName){
     let accessDenied = ['Liam', 'Noah', 'William', 'James', 'Logan', 'Benjamin',
                        'Mason', 'Elijah', 'Oliver', 'Jacob', 'Daniel', 'Lucas'];
      switch (varName) {
         case (accessDenied.includes(varName) ? varName : null):
             return 'Access Denied!';
         default:
           return 'Access Allowed.';
       }
    }

    console.log(isAccessible('Liam'));

switch语句用于根据条件从许多代码块中选择要执行的一个

switch表达式中的值与提供的不同值进行比较 如果匹配，则执行与之相关的代码块 如果没有匹配，则执行默认块

语法:

switch(expression) {
  case x:
    // code block
    break;
  case y:
    // code block
    break;
  default:
    // code block
}

注意: 必须注意的是，如果省略了break语句，那么即使它们与switch表达式不匹配，下一个块也将被执行。因此，如果不想获得指定的行为，不要忘记在每个代码块的末尾添加break语句

一个实际的例子: 下面的代码基于一个整数(由'new Date().getDay()'提供)以字符串形式返回一周中的当前日期。

switch (new Date().getDay()) {
  case 0:
    day = "Sunday";
    break;
  case 1:
    day = "Monday";
    break;
  case 2:
     day = "Tuesday";
    break;
  case 3:
    day = "Wednesday";
    break;
  case 4:
    day = "Thursday";
    break;
  case 5:
    day = "Friday";
    break;
  case 6:
    day = "Saturday";
}

代码样本取自W3Schools

上述方法的问题在于，每次调用具有开关的函数时，都必须重复上述几种情况。更可靠的解决方案是使用地图或字典。

这里有一个例子:

// The Map, divided by concepts var dictionary = { timePeriod: { 'month': [1, 'monthly', 'mensal', 'mês'], 'twoMonths': [2, 'two months', '2 months', 'bimestral', 'bimestre'], 'trimester': [3, 'trimesterly', 'quarterly', 'trimestral'], 'semester': [4, 'semesterly', 'semestral', 'halfyearly'], 'year': [5, 'yearly', 'annual', 'ano'] }, distance: { 'km': [1, 'kms', 'kilometre', 'kilometers', 'kilometres'], 'mile': [2, 'mi', 'miles'], 'nordicMile': [3, 'Nordic mile', 'mil (10 km)', 'Scandinavian mile'] }, fuelAmount: { 'ltr': [1, 'l', 'litre', 'Litre', 'liter', 'Liter'], 'gal (imp)': [2, 'imp gallon', 'imperial gal', 'gal (UK)'], 'gal (US)': [3, 'US gallon', 'US gal'], 'kWh': [4, 'KWH'] } }; // This function maps every input to a certain defined value function mapUnit (concept, value) { for (var key in dictionary[concept]) { if (key === value || dictionary[concept][key].indexOf(value) !== -1) { return key } } throw Error('Uknown "'+value+'" for "'+concept+'"') } // You would use it simply like this mapUnit("fuelAmount", "ltr") // => ltr mapUnit("fuelAmount", "US gal") // => gal (US) mapUnit("fuelAmount", 3) // => gal (US) mapUnit("distance", "kilometre") // => km // Now you can use the switch statement safely without the need // to repeat the combinations every time you call the switch var foo = 'monthly' switch (mapUnit ('timePeriod', foo)) { case 'month': console.log('month') break case 'twoMonths': console.log('twoMonths') break case 'trimester': console.log('trimester') break case 'semester': console.log('semester') break case 'year': console.log('year') break default: throw Error('error') }

如果您的case条件很复杂，需要很多case值匹配，或者需要动态值匹配，那么最好将case匹配逻辑移动到处理程序子函数中。

在您的情况下，如果您有成千上万的用户名来匹配安全权限检查，这种方法是更清晰的选项，更可扩展，可以暴露高级多路分支逻辑，而不会陷入一长串case语句中。

    switch (varName)
    {     
      case checkPatternAdministrator(varName):
        alert('Hey');
        break;

      case checkPatternUserTypeA(varName):
        alert('Hey2');
        break;

      case checkPatternUserTypeB(varName):
        alert('Hey3');
        break;

      default:
        alert('Default case');
        break;
    }

 function checkPatternAdministrator(varName) {
    // Logic to check Names against list, account permissions etc.
    // return the varName if a match is found, or blank string if not
    var matchedAdministratorName = varName;

    return matchedAdministratorName;
  }

你可以使用'in'操作符… 它依赖于对象/散列调用，所以它的速度和JavaScript一样快。

// Assuming you have defined functions f(), g(a) and h(a,b)
// somewhere in your code,
// you can define them inside the object, but...
// the code becomes hard to read. I prefer it this way.

o = { f1:f, f2:g, f3:h };

// If you use "STATIC" code can do:
o['f3']( p1, p2 )

// If your code is someway "DYNAMIC", to prevent false invocations
// m brings the function/method to be invoked (f1, f2, f3)
// and you can rely on arguments[] to solve any parameter problems.
if ( m in o ) o[m]()