我有以下把手结构:
├── gulpfile.js
└── source/
└── templates/
├── view1/
│ └── template11.handlebars
└── view2/
└── template21.handlebars
得到这个:
└── target/
└── js/
├── view1.min.js
└── view2.min.js
问题是如何创建实际缩小的预编译模板。现在我只是打开预编译的js。
我的 gruntfile.js 是:
const pump = require( 'pump' )
const rename = require( 'gulp-rename' )
const handlebars = require( 'gulp-handlebars' )
gulp.task( 'build-templates', ( done ) => {
const views = [
'view1',
'view2'
]
let pipe = []
views.forEach( ( view ) => {
pipe.push( gulp.src( 'source/templates/' + view + '/**/*' ) )
pipe.push( handlebars() )
pipe.push( rename( view +'.templates.min.js' ) )
// pipe.push( uglify() ) <-- this gives me the error:
// [13:40:38] GulpUglifyError: unable to minify JavaScript
// Caused by: SyntaxError: Unexpected token: punc (:) (line: 1, col: 11, pos: 11)"
pipe.push( gulp.dest( 'target/js' ) )
} )
pump( pipe, done )
} )
我使用pump
只是让node.js知道如果进程在处理管道时产生错误,就必须关闭源代码。
谢谢! :)
答案 0 :(得分:2)
我没有意识到我需要将编译后的代码作为参数包装到Handlebars.template()
。在gulp-handlebars
文档中明确指出。 :(所以结果不是一个有效的js代码,uglify无法处理它。解决方案是:
const pump = require( 'pump' )
const concat = require( 'gulp-concat' )
const wrap = require( 'gulp-wrap' )
const declare = require( 'gulp-declare' )
const handlebars = require( 'gulp-handlebars' )
const uglify = require( 'gulp-uglify' )
gulp.task( 'build-templates', ( done ) => {
const views = [
'view1',
'view2'
]
let pipe = []
views.forEach( ( view ) => {
pipe.push( gulp.src( 'source/templates/' + view + '/**/*' ) )
pipe.push( handlebars() )
pipe.push( wrap( 'Handlebars.template(<%= contents %>)' ) ) // <-- this is the key
pipe.push( declare( {
namespace: 'MyApp.templates', // <-- get easy access to templates
noRedeclare: true, // <-- Avoid duplicate declarations
} ) )
pipe.push( concat( view + '.templates.js' ) ) // <-- use concat instead of rename to concatenate several templates
pipe.push( uglify() ) // <-- done
pipe.push( gulp.dest( 'target/js' ) )
} )
pump( pipe, done )
} )