Problem in the execution of the task with Gulp & SASS?

Question

I am creating a task that compiles, renames and simplifies files .scss for .css. My file structure is like this:

assets/
|__ css/
|__ sass/
|   |__ uma-pasta/
|   |   |__ alguns.scss
|   |   |__ arquivos.scss
|   |   |__ vão.scss
|   |   |__ aqui.scss
|   |__ outra-pasta/
|       |__ alguns.scss
|       |__ outros.scss
|       |__ arquivos.scss
|       |__ vão.scss
|       |__ aqui.scss
|__ _variables.scss

But when I perform the following task:

gulp.task('sass', function() {
    gulp.src(['!assets/sass/outra-pasta/**/*.scss', 'assets/sass/**/*.scss'])
        .pipe(sass().on('error', sass.logError))
        .pipe(rename({ suffix: '.min' }))
        .pipe(minify())
        .pipe(gulp.dest(PATH + 'assets/css'))
});

The variables located in the assets/sass/_variables.scss are not concatenated with the other files, and then I get the following error:

Error in plugin 'sass'
Message:
    assets/sass/uma-pasta/arquivos.scss
Error: Undefined variable: "$variable".
    on line 3 of stdin
>>  width: $variable;
--------^

And the file assets/sass/_variables.scss is that way:

$variable: 20px;

Answer 1

I confess I never used gulp, but if the logic is the same as grunt, I believe you should start by inserting the variable file before inserting the other files. Try the following:

gulp.src([
    'assets/sass/variables.scss', 
    'assets/sass/**/*.scss'
])

If you have other files still in the root folder, just use something like this:

gulp.src([
    'assets/sass/variables.scss', 
    'assets/sass/*.scss', 
    'assets/sass/**/*.scss'
])

So variables will exist for all files.

---

Edited:

Another observation, I don’t know if it was your mistake when passing the question to the OS or if it really is the error. But the log says that you did not find the variable $minhavariavel and the name you gave next was: $variable. Make sure there is no syntax error with the variable names.