typescript 我如何在打字稿构造函数之外正确使用 angular $timeout 服务？
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How do i correctly use angular $timeout service outside of a typescript constructor?
提问by XT_Nova
I'm trying to figure out how to correctly use the angular $timeout service with TypeScript - So that i can show a splash-screen-like-behavior when someone for example clicks on a "Log In"-button.
我试图弄清楚如何正确使用带有 TypeScript 的 angular $timeout 服务 - 这样我就可以在某人单击“登录”按钮时显示类似闪屏的行为。
interface INavigationController {
username: string;
password: string;
loggedIn: boolean;
ready: boolean;
signIn(): void;
signOut(): void;
}
class NavigationController implements INavigationController {
username: string;
password: string;
loggedIn: boolean;
ready: boolean;
static $inject = ['$timeout'];
constructor(private $timeout: ng.ITimeoutService) {
var vm = this;
vm.username = this.username;
vm.password = this.password;
vm.ready = true;
// The line below work just fine from within here - but i want the behavior
// when someone clicks a button, not directly like this.
// Remove vm.ready = true above, before...
//$timeout(function () { vm.ready = true;}, 2200);}
signIn() {
if (!this.username || !this.password) {
alert("You need to frickin enter som details, man!");
} else {
this.ready = false;
this.$timeout(function () {
this.ready = true;
}, 2200);
}
}
I have also tried to set the invokeApply?:booleanto true:
我还尝试将 设置invokeApply?:boolean为 true：
this.$timeout(function () {
this.ready = true;
}, 2200, true);
- Without any success...
- 没有任何成功...
If i console.log(this.ready);from within the this.$timeout(function()...it will show me trueafter 2200ms, but it seems that the $apply()doesn't apply?
如果我console.log(this.ready);从内部this.$timeout(function()...它会true在 2200 毫秒后显示给我，但似乎$apply()不适用？
Would appreciate if someone could enlighten me in how to use the $timeout and other services like it, correctly - Preferably using the style-guide i'm aiming for i.e. using "controllerAs with vm" - John Papa
如果有人能启发我正确地使用 $timeout 和其他类似的服务，我将不胜感激 - 最好使用我的目标风格指南，即使用“controllerAs with vm” - John Papa
回答by lujcon
this.$timeout(()=> {
this.ready = true;
}, 2200);
if you use function(){} pattern you don't have access to this(thisis not the one you expect)
如果您使用 function(){} 模式，则您无权访问this（this不是您期望的模式）
回答by Peter Ashwell
The 'this' inside the timeout is the this of the timeout function. You need to define a variable outside to reference that scope (usually called 'that') and refer to it inside the timeout:
超时内的'this'是超时函数的this。您需要在外部定义一个变量来引用该范围（通常称为“那个”）并在超时内引用它：
var that = this;
$timeout(function() {
that.value ='eggs';
}, 2200);
回答by Hall Wong
try this, mate
试试这个，伙计
var obj = this;
this.$timeout(function () {
obj.ready = true;
}, 2200);
