缩进Sass语法不适用于node-sass和gulp-sass [英] Indented Sass syntax not working with node-sass and gulp-sass

问题描述

Libsass 2.0为libsass用户带来了缩进语法，但到目前为止，我无法使用 node-sass 和 gulp -sass 。我有所有最新版本：

node-sass ：0.93

gulp-sass ：0.7.2

gulp ：3.8.2

此设置编译。 scss 文件，甚至 .sass 文件，但不会编译缩进语法。有没有人成功编译缩进语法与节点sass和gulp？

我 gulpfile.js

  var gulp = require（'gulp'）; 
 var sass = require（'gulp-sass'）; 
 
 gulp.task（'sass'，function（）{
 return gulp.src（'./ sites / all / themes / nsfvb / sass / screen.sass'）
 .pipe（sass（{
 includePaths：require（'node-neat'）。includePaths，
 errLogToConsole：true 
} 
））
 .pipe .dest（'./ sites / all / themes / nsfvb / css'））; 
}）; 
 
 gulp.task（'watch'，function（）{
 gulp.watch（'./ sites / all / themes / nsfvb / sass / *。sass'，['sass' ]）; 
}）; 
 
 gulp.task（'default'，['sass'，'watch']）; 
  

运行默认任务时出错 / p>

错误：顶级表达式无效

解决方案

更新答案：

如果要使用缩进语法文件，使用sass（{indentedSyntax：true}）。

  sass }）
  

过期的答案

这里： https://github.com/dlmanning/gulp-sass/issues / 55＃issuecomment-50882250

使用默认设置编译sass文件不起作用。但是有一个解决方法。如果你传递sourceComments：'normal'作为参数，编译工作。原因是，有一个奇怪的条件，它改变了文件的处理方式： https://github.com/dlmanning/gulp-sass/blob/master/index.js#L23-L27

在此处找到代码示例： https：/ /github.com/chrisdl/gulp-libsass-example/blob/master/gulpfile.js

  var gulp = require（'gulp'）; 
 var sass = require（'gulp-sass'）; 
 
 //在终端中运行> gulp sass。 
 gulp.task（'sass'，function（）{
 gulp.src（'./ sass / main.sass'）
 .pipe（sass（{sourceComments：'normal'} ））
 .pipe（gulp.dest（'./ css'））; 
}）; 
  

注意

使用此代码段查看以下引用和问题。

使用此解决方法将导致对文件内容的任何更改被丢弃的gulp管道（例如早期的插件） - JMM

https://github.com/dlmanning/gulp-sass/issues/158

Libsass 2.0 brought the indented syntax to libsass users, but so far I've been unable to make it work with node-sass and gulp-sass. I have all of the latest versions:

node-sass: 0.93
gulp-sass: 0.7.2
gulp: 3.8.2

This setup compiles .scss files and even .sass files using the bracket syntax but it will not compile the indented syntax. Has anyone successfully compiled the indented syntax with node-sass and gulp?

My gulpfile.js

var gulp = require('gulp');
var sass = require('gulp-sass');

gulp.task('sass', function() {
    return gulp.src('./sites/all/themes/nsfvb/sass/screen.sass')
        .pipe(sass({
            includePaths: require('node-neat').includePaths,
            errLogToConsole: true
        }
        ))
        .pipe(gulp.dest('./sites/all/themes/nsfvb/css'));
});

gulp.task('watch', function() {
    gulp.watch('./sites/all/themes/nsfvb/sass/*.sass', ['sass']);
});

gulp.task('default', ['sass', 'watch']);

Error when running the default task

error: invalid top-level expression

解决方案

Updated answer:

If you want to use the indented syntax (.sass) as the top level file, use sass({indentedSyntax: true}).

sass({indentedSyntax: true})

Outdated answer

Answer found here: https://github.com/dlmanning/gulp-sass/issues/55#issuecomment-50882250

With default settings compiling sass files doesn't work. However there is a workaround. If you pass sourceComments: 'normal' as parameter the compilation work. The reason for this is that there is a strange condition which change how file are handled: https://github.com/dlmanning/gulp-sass/blob/master/index.js#L23-L27

Code example found here: https://github.com/chrisdl/gulp-libsass-example/blob/master/gulpfile.js

var gulp = require('gulp');
var sass = require('gulp-sass');

// Run with "> gulp sass" in terminal.
gulp.task('sass', function () {
gulp.src('./sass/main.sass')
    .pipe(sass({sourceComments: 'normal'}))
    .pipe(gulp.dest('./css'));
});

Notice

If you run into issues using this snippet take a look at the following quote and issue.

Using this workaround will result in any changes to the file content from earlier in the gulp pipeline (e.g. earlier plugins) being discarded - JMM

https://github.com/dlmanning/gulp-sass/issues/158

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