如何在gulp任务中仅排除.min.js文件，同时保留在同一目录中？

Question

I am trying to set up a build system for my front end work though I am running into a problem where it loops processing files over and over again. This is a problem with my js processing since I am not sure how to exclude just the files with .min as a suffix.

The task goes as follows

return gulp.src(["!dev/js/*.min.js", "dev/js/*.js"])
        .pipe(plumber())
        .pipe(browserify())
        .pipe(smaps.init())
            .pipe(uglyify({preserveComments: "license"}))
        .pipe(smaps.write())
        .pipe(rename({suffix: ".min"}))
        .pipe(gulp.dest(output_dir));

Though what I have found is that it still targets the .min.js files since they are also seen as .js files. I have messed around with a few different configurations of these wildcards but I keep ending up with the task looping creating example.min.js then example.min.min.js then example.min.min.min.js etc.

So, how can I just process files that do not include the .min prefix?

Answer 1

您可以使用否定模式排除.min.js文件。

gulp.src(['dev/js/*.js', '!dev/js/*.min.js'])

Answer 2

如果您只想使用一个字符串来执行此操作，则可以使用：

gulp.src(["dev/js/!(*.min)*.js"])

Answer 3

In your gulp command.

const
    gulpIgnore = require( 'gulp-ignore' ),
    uglify = require( 'gulp-uglify' ),
    jshint = require( 'gulp-jshint' ),
    condition = './**/*.min.js';

gulp.task( 'task', function() {
    gulp.src( './**/*.js' )
        .pipe( jshint() )
        .pipe( gulpIgnore.exclude( condition ) )
        .pipe( uglify() )
        .pipe( gulp.dest( './dist/' ) );
} );

In the condition you specify your directory. ../ to go backward, /dir to go forwards