防止在本地计算机上对代码进行缩小

Question

我用[RDash angular dashboard][1]启动了一个项目，该项目使用了gump。这是我第一次使用gump，问题是当我在本地工作时，我无法调试，因为它缩小了css/js/html文件。

在本地工作时，如何防止吞咽变小？

以下是我在gulpfile.js:上的内容

var gulp = require('gulp'),
usemin = require('gulp-usemin'),
wrap = require('gulp-wrap'),
connect = require('gulp-connect'),
watch = require('gulp-watch'),
minifyCss = require('gulp-minify-css'),
minifyJs = require('gulp-uglify'),
concat = require('gulp-concat'),
less = require('gulp-less'),
rename = require('gulp-rename'),
minifyHTML = require('gulp-minify-html');
var paths = {
scripts: 'src/js/**/*.*',
styles: 'src/less/**/*.*',
images: 'src/img/**/*.*',
templates: 'src/templates/**/*.html',
index: 'src/index.html',
bower_fonts: 'src/components/**/*.{ttf,woff,eof,svg}',
};
/**
 * Handle bower components from index
   */
   gulp.task('usemin', function() {
return gulp.src(paths.index)
    .pipe(usemin({
        js: [minifyJs(), 'concat'],
        css: [minifyCss({keepSpecialComments: 0}), 'concat'],
    }))
    .pipe(gulp.dest('dist/'));
});
/**
 * Copy assets
   */
gulp.task('build-assets', ['copy-bower_fonts']);
gulp.task('copy-bower_fonts', function() {
return gulp.src(paths.bower_fonts)
    .pipe(rename({
        dirname: '/fonts'
    }))
    .pipe(gulp.dest('dist/lib'));
});
/**
 * Handle custom files
 */
gulp.task('build-custom', ['custom-images', 'custom-js', 'custom-less',         'custom-templates']);
gulp.task('custom-images', function() {
return gulp.src(paths.images)
    .pipe(gulp.dest('dist/img'));
});
gulp.task('custom-js', function() {
return gulp.src(paths.scripts)
    .pipe(minifyJs())
    .pipe(concat('dashboard.min.js'))
    .pipe(gulp.dest('dist/js'));
});
gulp.task('custom-less', function() {
return gulp.src(paths.styles)
    .pipe(less())
    .pipe(gulp.dest('dist/css'));
});
gulp.task('custom-templates', function() {
return gulp.src(paths.templates)
    .pipe(minifyHTML())
    .pipe(gulp.dest('dist/templates'));
});
/**
 * Watch custom files
 */
gulp.task('watch', function() {
gulp.watch([paths.images], ['custom-images']);
gulp.watch([paths.styles], ['custom-less']);
gulp.watch([paths.scripts], ['custom-js']);
gulp.watch([paths.templates], ['custom-templates']);
gulp.watch([paths.index], ['usemin']);
});
/**
 * Live reload server
 */
gulp.task('webserver', function() {
connect.server({
    root: 'dist',
    livereload: true,
    port: 8888
});
});
gulp.task('livereload', function() {
gulp.src(['dist/**/*.*'])
    .pipe(watch())
    .pipe(connect.reload());
});
/**
 * Gulp tasks
 */
gulp.task('build', ['usemin', 'build-assets', 'build-custom']);
gulp.task('default', ['build', 'webserver', 'livereload', 'watch']);

Answer 1

您应该添加gullow minify并按如下方式使用它：

var minify = require('gulp-minify');
gulp.task('compress', function() {
  gulp.src('lib/*.js')
    .pipe(minify({
        exclude: ['tasks'],
        ignoreFiles: ['.combo.js', '-min.js']
    }))
    .pipe(gulp.dest('dist'))
});

将不想缩小的文件传递给exclude。

在您的代码中，您已经指定要缩小此处的所有文件：

  gulp.task('usemin', function() {
   return gulp.src(paths.index)
    .pipe(usemin({
        js: [minifyJs(), 'concat'],
        css: [minifyCss({keepSpecialComments: 0}), 'concat'],
    }))
    .pipe(gulp.dest('dist/'));
});

在线

.pipe(usemin({..}) 

只是不要使用这个usemin任务，你应该很好