使用模块名称而不是文件名加载模块
我在两个单独的 js 文件中有两个“定义”。
def1.js 和 def2.js
define("mydefname1",["file1",...]});
define("mydefname2",["file2",....]});
我有另一个 require 语句，我检查这两个定义是否已加载。
require(['def1','def2'], function(){alert('loaded')});
这工作正常..
但如果我尝试
require(['mydefname1','mydefname2'], function(){alert('loaded')});,
它就不起作用。
有没有办法我实际上可以使用 mydefname1 和 mydefname2.. 即模块名称来加载它们，而不是文件名？
I have two 'define' in two separate js files.
def1.js and def2.js
define("mydefname1",["file1",...]});
define("mydefname2",["file2",....]});
I have another require statement where i check if the two definetions are loaded.
require(['def1','def2'], function(){alert('loaded')});
this works fine..
but if I try
require(['mydefname1','mydefname2'], function(){alert('loaded')});,
it does not work.
Is there a way I could actually use mydefname1 and mydefname2.. i.e. the module name to load them, and not the file name?
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据我所知，这是不可能的。我通常做的就是格式化我的文件名以包含我的模块名称，这样我就可以这样包含它。
对于元类名称 Overlay，文件名将为 class.Overlay.js，我的 require/include 函数将采用我的模块名称并从中构建文件名。
As far as I know, that's not possible. What I usually do is format my file names to include my module name, so I can include it that way.
For a metaclass name Overlay, the file name would be class.Overlay.js and my require/include functions would take my module name and build the file name from it.
据我了解，RequireJS 建议您避免手动分配模块名称 - 请参阅此处： http://requirejs .org/docs/api.html#modulename
FWIW，您是否使用过“path”配置选项？我认为您不能用它来完全完成您想要做的事情，但它可能提供一种可接受的替代方法。请参阅此处：http://requirejs.org/docs/api.html#config
希望有帮助！
As I understand it, RequireJS recommends that you avoid manually assigning module names - see here: http://requirejs.org/docs/api.html#modulename
FWIW, have you played around with the "path" configuration option? I don't think you can use it to do exactly what you are looking to do but it may offer an acceptable alternate approach. See here: http://requirejs.org/docs/api.html#config
Hope it helps!