如何使用RequireJS构建多个页面

Question

如何使用RequireJS构建多个页面？就像下面的示例一样,声明app.js中的每个类都是正确的做法吗？有没有每个html文件声明<script data-main="src/main" src="src/require.js"></script>？

我想避免的是当用户到达网站的第一页时加载所有脚本.

main.js定义所有外部依赖项:

require(
    {
        baseUrl:'/src'
    },
    [
        "require",
        "order!http://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js",
        "order!http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.13/jquery-ui.min.js",
        "order!http://ajax.cdnjs.com/ajax/libs/underscore.js/1.1.6/underscore-min.js",
        "order!http://ajax.cdnjs.com/ajax/libs/backbone.js/0.3.3/backbone-min.js"
    ], 
    function (require) {
        require(["app"], function (app) {
            app.start();
        });
    }
);

定义每个组件的app.js文件:

define([ "product/ProductSearchView",
         "product/ProductCollection"
         ], function (ProductSearchView,
                      ProductCollection) {
         return {
             start: function() {
                  var products = new ProductCollection();
                  var searchView = new ProductSearchView({ collection: products });
                  products.fetch();
                  return {};
             }
        }
});

Answer 1

您可以在现有模块中要求文件.因此,当有人点击链接时,您可以触发执行以下操作的功能:

// If you have a require in your other module
// The other module will execute its own logic
require(["module/one"], function(One) {
    $("a").click(function() {
        require(["new/module"]);
    });
});

// If you have a define in your other module
// You will need to add the variable to the require
// so you can access its methods and properties
require(["module/one"], function(One) {
    $("a").click(function() {
        require(["new/module"], function(NewModule) {
            NewModule.doSomething();
        });
    });
});