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？是否每个html文件都声明< 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 {};
}
}
});
解决方法
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();
});
});
});
