Javascript开关语句?

Javascript开关语句?,javascript,Javascript,我有如下代码: for (var i = 0; i < $scope.Option.length; i++) { var option = $scope.Option[i].Code; if (option == "A") { $scope.aSelected = true; break; } } for (var i = 0; i < $scope.Option.length; i++) { var option

我有如下代码:

for (var i = 0; i < $scope.Option.length; i++) {
    var option = $scope.Option[i].Code;

    if (option == "A") {
        $scope.aSelected = true;
        break;
    }
}

for (var i = 0; i < $scope.Option.length; i++) {

    var option = $scope.Option[i].Code;

    if (option == "B") {
        $scope.bSelected = true;
        break;
    }
}
for (var i = 0; i < $scope.Option.length; i++) {

    var option = $scope.Option[i].Code;

    switch (option) {
        case "A":
            $scope.aSelected = true;
            break;
        case "B":
            $scope.bSelected = true;
            break;
        default:
            console.log('unrecognized option');
    }
}
for(变量i=0;i<$scope.Option.length;i++){
var option=$scope.option[i]。代码;
如果(选项==“A”){
$scope.aSelected=true;
打破
}
}
对于(变量i=0;i<$scope.Option.length;i++){
var option=$scope.option[i]。代码;
如果(选项==“B”){
$scope.bSelected=true;
打破
}
}
是否可以在下面的switch语句中对此进行更正:

for (var i = 0; i < $scope.Option.length; i++) {
    var option = $scope.Option[i].Code;

    if (option == "A") {
        $scope.aSelected = true;
        break;
    }
}

for (var i = 0; i < $scope.Option.length; i++) {

    var option = $scope.Option[i].Code;

    if (option == "B") {
        $scope.bSelected = true;
        break;
    }
}
for (var i = 0; i < $scope.Option.length; i++) {

    var option = $scope.Option[i].Code;

    switch (option) {
        case "A":
            $scope.aSelected = true;
            break;
        case "B":
            $scope.bSelected = true;
            break;
        default:
            console.log('unrecognized option');
    }
}
for(变量i=0;i<$scope.Option.length;i++){
var option=$scope.option[i]。代码;
开关(选件){
案例“A”:
$scope.aSelected=true;
打破
案例“B”:
$scope.bSelected=true;
打破
违约:
log(“无法识别的选项”);
}
}

在切换情况下,这实际上是不正确的,因为第一个选项可能是A,它将中断循环,然后,例如,如果“B”是集合后面位置的选项,它将永远不会得到bSelected=true

这是正确的,因为开关内部的断路器会断开,而不是整个FOR循环,因此它也会检查B。

“这在开关情况下实际上不正确吗”-您测试过吗?它是否有你推测的行为?@Quentin-很好的观点-应该做一个快速测试-道歉