我尝试使用TypeScript和Angular编写动态模板,但出于某种原因,' this'关键字始终为null,因此我无法访问我的私有成员$ compile。有任何想法吗?非常感谢! : - )
指令:
namespace ROD.Features.Player {
"use strict";
export class VideoDirective implements ng.IDirective {
public restrict: string = "E";
public replace: boolean = true;
public scope = {
content: "="
};
constructor(private $compile: ng.ICompileService) {
}
public link(element: JQuery, scope: ng.IScope): any {
const youtubeTemplate = "<p>Youtube</p>";
const vimeoTemplate = "<p>Vimeo</p>";
var linkFn = this.$compile(youtubeTemplate);
const content: any = linkFn(scope);
element.append(content);
}
}
}
App.ts:
namespace ROD {
"use strict";
angular.module("rodApp", [])
.service("Settings", [() => new Settings.DevelopmentSettings()])
.service("RedditService", [
"$http", "Settings",
($http: ng.IHttpService, settings: Settings.ISettings) => new Services.Reddit.RedditService($http, settings.sourceUrl),
])
.directive("videoItem", ["$compile",
($compile: ng.ICompileService) => new Features.Player.VideoDirective($compile)])
.controller("PlayerController", [
"$scope", "RedditService",
($scope: any, redditService: Services.Reddit.IRedditService) => new Features.Player.PlayerController($scope, redditService),
]);
}
答案 0 :(得分:3)
我似乎使用了错误的语法来处理链接功能。这是正确的实施:
public link = (element: JQuery, scope: ng.IScope): any => {
const youtubeTemplate = "<p>Youtube</p>";
const vimeoTemplate = "<p>Vimeo</p>";
var linkFn = this.$compile(youtubeTemplate);
const content: any = linkFn(scope);
element.append(content);
}
任何人都能解释为什么会这样吗? : - )
答案 1 :(得分:1)
你需要:
.directive("videoItem", ["$compile",
function ($compile) { return () => new ROD.Features.Player.VideoDirective($compile); }])
而不是
.directive("videoItem", ["$compile",
function ($compile) { return new ROD.Features.Player.VideoDirective($compile); }])
在app.js
中。这个问题的解释如下:http://www.michaelbromley.co.uk/blog/350/exploring-es6-classes-in-angularjs-1-x#_section-directives问题的要点是当角度调用link
起作用时,不会保留此上下文。
如果您想更深入地了解这个问题,只需使用angular.js
代替angular.min.js
,看看调用堆栈是什么样的。