Question

有几个像这样的问题，但我相信我的情况略有不同，我似乎无法弄明白。

给出这个伪代码

1.) BaseFolder/*.js except the app.js file
2.) BaseFolder/app/model/*js
3.) BaseFolder/app/store/*.js
4.) BaseFolder/app/view/*.js except Viewport.js
5.) BaseFolder/app/view/Viewport.js
6.) BaseFolder/app/controller/*.js
7.) BaseFolder/app.js

我的问题是我否定了app.js文件，然后我想在最后重新添加它。同样处理Viewport.js文件。

任何想法如何解决这个问题？

这是我尝试过的很多事情之一：

var senchaFiles = [
            baseFolderPath + '/*.js',
            '!' + baseFolderPath + '/app.js',
            baseFolderPath + '/app/model/*.js',
            baseFolderPath + '/app/store/*.js',
            baseFolderPath + '/app/view/*.js',
            '!' + baseFolderPath + '/app/view/Viewport.js',
            baseFolderPath + '/app/view/Viewport.js',
            baseFolderPath + '/app/controller/*.js',
            baseFolderPath + '/app.js'
        ];

return gulp.src(senchaFiles)
            .pipe(concat(folder + '.js'))
            // .pipe(sourcemaps.init())
            //.pipe(gulp.dest(JS_DIST_FOLDER))
            // .pipe(uglify())
            // .pipe(rename(folder + '.min.js'))
            // .pipe(sourcemaps.write())
            .pipe(gulp.dest(JS_DIST_FOLDER));
    });

运行此代码不会添加回app.js或{i} Viewport.js文件中。

Answer 1

Gulp使用node-glob syntax。以下代码可能是您正在寻找的内容。

var senchaFiles = [
  baseFolderPath + '/!(app)*.js', // all, but app.js
  baseFolderPath + '/app/model/*.js',
  baseFolderPath + '/app/store/*.js',
  baseFolderPath + '/app/view/!(Viewport)*.js', // all, but Viewport.js
  baseFolderPath + '/app/view/Viewport.js',
  baseFolderPath + '/app/controller/*.js',
  baseFolderPath + '/app.js'
];

Gulpjs否定文件然后重新添加它

1 个答案: