gulp undefined不是一个功能

Question

这里我是node，angular，gulp的新手。我正在使用gulp生成minify文件所有的东西，而我在运行gulp任务时给了我错误

下面是我的gulpfile.js代码

gulp.task('dev', function(){
var builtPartialDev = gulp.src(paths.partials)
    .pipe(plugins.htmlhint({'doctype-first':false}))
    .pipe(plugins.htmlhint.reporter())
    .pipe(gulp.dest(paths.distDev));

var buitStyleDev = gulp.src(paths.styles)
    .pipe(plugins.less())
    .pipe(gulp.dest(paths.distDev));

var builtAppScriptsDev = gulp.src(paths.scripts)
    .pipe(plugins.jshint())
    .pipe(plugins.jshint.reporter('jshint-stylish'))
    .pipe(gulp.dest(paths.distDev))
    .pipe(plugins.angularFilesort());

var builtVendorScriptsDev = gulp.src(bowerFiles())
    .pipe(gulp.dest(paths.distDevBowerComponents))
    .pipe(plugins.order(['jquery.js', 'angular.js', 'angular-ui.js', 'ui-bootstrap.js', 'ui-bootstrap-tpls.js']));


var builtIndexDev = gulp.src(paths.index)
    .pipe(plugins.htmlhint())
    .pipe(plugins.htmlhint.reporter())
    .pipe(gulp.dest(path.distDev))
    .pipe(plugins.inject(builtVendorScriptsDev, {relative:true, name:'bower'}))
    .pipe(plugins.inject(builtAppScriptsDev, {relative:true}))
    .pipe(plugins.inject(buitStyleDev, {relative:true}))
    .pipe(gulp.dest(paths.distDev));

return es.merge(builtIndexDev, builtPartialDev) 

});

Answer 1

你需要在gulpfile中输入gulp。将以下行添加到您的文件中：

var gulp = require('gulp')

您可能还需要为项目安装gulp，您可以使用npm install --save gulp。

如果您没有npm，我建议您研究如何安装和使用它。