我有一个关于gulp的问题。我决定使用gulp自动化源代码和样式编译成单个文件的过程。但是,它不想覆盖由我的项目中所有.js文件创建的application.js文件。奇怪的是,它实际上覆盖了从项目中所有.less文件生成的已编译css文件。以下是我项目文件结构的样子:
如您所见,gulp任务
我尝试在
那么,有什么建议可以解决问题吗?
.
├── gulpfile.js
└── apps
├── base
├── controllers
├── controllerBaseOne.js
└── controllerBaseTwo.js
├── directives
├── directiveBaseOne.js
└── directiveBaseTwo.js
├── services
└── serviceBaseOne.js
└── styles
└── styleBase.less
├── header.html
└── index.html
├── services
├── compilation
├── application.js
└── application.css
├── controllers
├── controllerServicesOne.js
├── controllerServicesTwo.js
└── controllerServicesThree.js
├── directives
├── directiveServicesOne.js
├── directiveServicesTwo.js
└── directiveServicesThree.js
├── services
├── serviceServicesOne.js
└── serviceServicesTwo.js
└── styles
├── styleServicesOne.less
├── styleServicesTwo.less
└── styleServicesThree.less
├── header.html
└── index.html
├── appMain.js
└── config.json
现在我的gulpfile.js
的样子是这样的:
var gulp = require( "gulp" );
var gulpif = require( "gulp-if" );
var concat = require( "gulp-concat" );
var uglify = require( "gulp-uglify" );
var less = require( "gulp-less" );
var cleanCSS = require( "gulp-clean-css" );
// application components and paths:
var compileMinify = false;
var basePath = process.cwd() + "apps/base";
var webPath = process.cwd() + "apps/services";
var compilationPath = "compilation";
var appCompiledFileName = "application";
var stylePaths = [
basePath + "/styles/**/*.less",
webPath + "/styles/**/*.less"
];
var sourcePaths = [
basePath + "/**/*.js",
webPath + "/**/*.js"
];
gulp.task( "services-source", function() {
return gulp.src( sourcePaths )
.pipe( concat( appCompiledFileName + ".js" ) )
.pipe( gulpif( compileMinify, uglify() ) )
.pipe( gulp.dest( compilationPath, { cwd: webPath } ) );
} );
gulp.task( "services-styles", function() {
return gulp.src( stylePaths )
.pipe( concat( appCompiledFileName + ".less" ) )
.pipe( less() )
.pipe( gulpif( compileMinify, cleanCSS( { debug: true }, function( details ) {
console.log( details.name + " original size: " + details.stats.originalSize );
console.log( details.name + " minified size: " + details.stats.minifiedSize );
} ) ) )
.pipe( gulp.dest( compilationPath, { cwd: webPath } ) );
} );
gulp.task( "services", [ "services-source", "services-styles" ], function() {
gulp.watch( sourcePaths, [ "services-source" ] );
gulp.watch( stylePaths, [ "services-styles" ] );
} );
如您所见,gulp任务
services-source
正在遍历apps
文件夹及其子文件夹中的每个.js
文件,并将它们连接成一个单独的文件,应该放置在compilation
文件夹中。在services-styles
任务中也是这样做的,只是进行了一些较少的转换。还有检查是否对样式和源代码进行缩小处理,但默认情况下已禁用。我尝试在
services-source
任务末尾添加参数来覆盖目标文件,像这样:overwrite: true
,但似乎没有发生任何事情。当我运行gulpfile.js
时,application.js
文件会变得越来越大,而且似乎无法覆盖它。那么,有什么建议可以解决问题吗?
"!" + webPath + "/" + compilationPath + "/**/*.js"
非常感谢! :) - GBak