Question

我有一个导出ParentModule的文件及其导入模块的数组

export const PARENT_MODULES = [
    module1,
    ...
    someModule,
    ...
    moduleN
];

@NgModule({ imports: [...PARENT_MODULES] })
export class ParentModule{ }

除了一次导入（ModifiedParentModule而不是ParentModule）之外，我想创建与anotherModule相同的someModule。如果没有再次输入整个导入列表，有没有办法做到这一点？

我试过了：

export const replaceModules = module => 
    module.name === someModule.name ? anotherModule.name : module;

export const MODIFIED_PARENT_MODULES = PARENT_MODULES.map(replaceModules)

@NgModule({ imports: [...MODIFIED_PARENT_MODULES ]})
export class ModifiedParentModule{ }

但是它给出了错误

ERROR in TypeError: Cannot use 'in' operator to search for 'providers' in null
at C:\app-ext\node_modules\@angular\compiler-cli\src\ngtools_impl.js:111:73
at Array.filter (native)
at _extractLazyRoutesFromStaticModule (C:\app-ext\node_modules\@angular\compiler-cli\src\ngtools_impl.js:111:29)
at includeLazyRouteAndSubRoutes (C:\app-ext\node_modules\@angular\compiler-cli\src\ngtools_impl.js:66:25)
at Array.reduce (native)
at Object.listLazyRoutesOfModule (C:\app-ext\node_modules\@angular\compiler-cli\src\ngtools_impl.js:54:36)
at Function.NgTools_InternalApi_NG_2.listLazyRoutes (C:\app-ext\node_modules\@angular\compiler-cli\src\ngtools_api.js:91:39)
at AotPlugin._getLazyRoutesFromNgtools (C:\app-ext\node_modules\@ngtools\webpack\src\plugin.js:207:44)
at _donePromise.Promise.resolve.then.then.then.then.then (C:\app-ext\node_modules\@ngtools\webpack\src\plugin.js:443:24)
at process._tickCallback (internal/process/next_tick.js:103:7)

Answer 1

添加导出所有必需共享模块的BaseSharedParentModule

@NgModule({ 
imports: [...PARENT_MODULES],
exports: [...PARENT_MODULES]

 })
export class BaseSharedParentModule { }

ParentModule和ModiefiedParentModule都导入容器模块，如下所示

@NgModule({ 
imports: [BaseSharedParentModule],
 })
export class ParentModule{ }



@NgModule({ 
imports: [BaseSharedParentModule],
 })
export class ModifiedParentModule{ }

Angular - 导入已修改的模块列表

1 个答案: