Question

我分叉了这个存储库：https://github.com/angular-ui/ui-select并在此处进行了更改：https://github.com/zhen-w/ui-select。问题是，我对angular-ui/ui-select的源代码进行了一些更改，但是当我运行命令gulp以尝试生成我的文件的缩小版本时，我的所有更改似乎都被恢复并且我的缩小生成的文件没有我想要的新更改。以下是gulpfile.js：

的内容

var fs = require('fs');
var gulp = require('gulp');
var karma = require('karma').server;
var concat = require('gulp-concat');
var jshint = require('gulp-jshint');
var header = require('gulp-header');
var footer = require('gulp-footer');
var rename = require('gulp-rename');
var es = require('event-stream');
var del = require('del');
var uglify = require('gulp-uglify');
var minifyHtml = require('gulp-minify-html');
var minifyCSS = require('gulp-minify-css');
var templateCache = require('gulp-angular-templatecache');
var gutil = require('gulp-util');
var plumber = require('gulp-plumber');//To prevent pipe breaking caused by errors at 'watch'

var config = {
  pkg : JSON.parse(fs.readFileSync('./package.json')),
  banner:
      '/*!\n' +
      ' * <%= pkg.name %>\n' +
      ' * <%= pkg.homepage %>\n' +
      ' * Version: <%= pkg.version %> - <%= timestamp %>\n' +
      ' * License: <%= pkg.license %>\n' +
      ' */\n\n\n'
};

gulp.task('default', ['build','test']);
gulp.task('build', ['scripts', 'styles']);
gulp.task('test', ['build', 'karma']);

gulp.task('watch', ['build','karma-watch'], function() {
  gulp.watch(['src/**/*.{js,html}'], ['build']);
});

gulp.task('clean', function(cb) {
  del(['dist'], cb);
});

gulp.task('scripts', ['clean'], function() {

  var buildTemplates = function () {
    return gulp.src('src/**/*.html')
      .pipe(minifyHtml({
             empty: true,
             spare: true,
             quotes: true
            }))
      .pipe(templateCache({module: 'ui.select'}));
  };

  var buildLib = function(){
    return gulp.src(['src/common.js','src/*.js'])
      .pipe(plumber({
        errorHandler: handleError
      }))
      .pipe(concat('select_without_templates.js'))
      .pipe(header('(function () { \n"use strict";\n'))
      .pipe(footer('\n}());'))
      .pipe(jshint())
      .pipe(jshint.reporter('jshint-stylish'))
      .pipe(jshint.reporter('fail'));
  };

  return es.merge(buildLib(), buildTemplates())
    .pipe(plumber({
      errorHandler: handleError
    }))
    .pipe(concat('select.js'))
    .pipe(header(config.banner, {
      timestamp: (new Date()).toISOString(), pkg: config.pkg
    }))
    .pipe(gulp.dest('dist'))
    .pipe(uglify({preserveComments: 'some'}))
    .pipe(rename({ext:'.min.js'}))
    .pipe(gulp.dest('dist'));

});

gulp.task('styles', ['clean'], function() {

  return gulp.src('src/common.css')
    .pipe(header(config.banner, {
      timestamp: (new Date()).toISOString(), pkg: config.pkg
    }))
    .pipe(rename('select.css'))
    .pipe(gulp.dest('dist'))
    .pipe(minifyCSS())
    .pipe(rename({ext:'.min.css'}))
    .pipe(gulp.dest('dist'));

});

gulp.task('karma', ['build'], function() {
  karma.start({configFile : __dirname +'/karma.conf.js', singleRun: true});
});

gulp.task('karma-watch', ['build'], function() {
  karma.start({configFile :  __dirname +'/karma.conf.js', singleRun: false});
});

var handleError = function (err) {
  console.log(err.toString());
  this.emit('end');
};

我不知道为什么我的更改会被还原，因为这是我自己的分叉存储库。任何帮助，将不胜感激。谢谢！

Answer 1

dist目录中的所有内容都是由gulp创建的 - 而不仅仅是缩小的文件。如果要进行更改，请编辑src目录下的相应文件。

Gulp正在恢复我的所有变化？

1 个答案: