抛出浏览器的同名多个源图？ （吞掉-sourcemaps）

Question

Here is the gulpfile:

widgets.forEach(function(widget){
        gulp.src('widgets/' + widget + "/Widget.es6")
            .pipe(sourcemaps.init())
            .pipe(babel({ modules: 'amd' }))
            //.pipe(uglify())
            .pipe(rename('Widget.js'))
            .pipe(sourcemaps.write('.'))
            .pipe(gulp.dest('widgets/' + widget + "/"));
    });

I have several "widgets" which I need to process. Each needs its own sourcemap. However, each widget has its own Widget.js file (inside its directory). This results in a Widget.js.map being created in each widget directory. This would be fine except that when I open up the google chrome developer and try to debug the original file by clicking in the transpiled one it is getting mixed up. I think it is using the wrong sourcemap because they all have the same name.

Looking for a solution to this problem (I need to name the widget js file Widget.js because of the framework I am working with)

Thanks in advance.

Answer 1

Adding a base to gulp.src fixed this issue for me.

For example gulp.src('widgets/' + widget + '/Widget.es6', { base: './' })