我有一个 gulpfile.js 有一些任务来编译客户端咖啡脚本和我在 sass 中的样式。咖啡脚本编译正确,但 sass 编译不正确。
这是我的 gulpfile.js -
var gulp = require('gulp'),
util = require('gulp-util'),
sass = require('gulp-sass'),
coffee = require('gulp-coffee');
var paths = {
scripts: {
src: 'src/coffee/**/*.coffee',
dest: 'public/javascripts'
},
styles: {
src: 'src/*.sass',
dest: 'public/stylesheets'
}
};
gulp.task('scripts', function() {
return gulp.src(paths.scripts.src)
.pipe(coffee())
.pipe(gulp.dest(paths.scripts.dest));
});
gulp.task('sass', function () {
return gulp.src(paths.styles.src)
.pipe(sass({errLogToConsole: true}))
.pipe(gulp.dest(paths.styles.dest));
});
gulp.task('watch', function () {
gulp.watch(paths.scripts.src, ['scripts']);
gulp.watch(paths.styles.src, ['styles']);
});
gulp.task('default', ['scripts', 'watch']);
它必须在我的“监视”任务中,但它缺少什么?
* 编辑 *
这是我得到的输出 -
[gulp] Using file exampledir/gulpfile.js
[gulp] Working directory changed to exampledir/examplesite
[gulp] Running 'scripts'...
[gulp] Running 'watch'...
[gulp] Finished 'watch' in 14 ms
[gulp] Finished 'scripts' in 106 ms
[gulp] Running 'default'...
[gulp] Finished 'default' in 10 μs