javascript - Using variables in Gulp for the destination file name? - Stack Overflow

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I am new to gulp and I am wondering if what I want to achieve is practical or possible.

My projects structure:

root
|
components
|   |
|   component_1
|   |   styles.scss
|   |   actions.js
|   |   template.html
|   |   ...
|   component_2
|   |   styles.scss
|   |   template.html
|   |   ...
|
public
    |
    assets
         |
         css (dest)
         |    component_1.css
         |    component_2.css
         |    ...
         js (dest)

Now what I want is that Gulp stores the compiled css files in the according css folder in public/assets but uses the name of folder where it found the scss file. Is that possible? Do I need to pipe that to a plugin? Thanks! PS i do realize I could achieve that by just renaming the scss, but that's what I'd like to avoid.

I am new to gulp and I am wondering if what I want to achieve is practical or possible.

My projects structure:

root
|
components
|   |
|   component_1
|   |   styles.scss
|   |   actions.js
|   |   template.html
|   |   ...
|   component_2
|   |   styles.scss
|   |   template.html
|   |   ...
|
public
    |
    assets
         |
         css (dest)
         |    component_1.css
         |    component_2.css
         |    ...
         js (dest)

Now what I want is that Gulp stores the compiled css files in the according css folder in public/assets but uses the name of folder where it found the scss file. Is that possible? Do I need to pipe that to a plugin? Thanks! PS i do realize I could achieve that by just renaming the scss, but that's what I'd like to avoid.

• javascript
• css
• sass
• gulp

Share Improve this question Follow asked Mar 19, 2014 at 16:51 Erik PöhlerErik Pöhler 75222 gold badges99 silver badges2222 bronze badges

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2 Answers 2

Sorted by: Reset to default 16

It wouldn't be too hard, depending on how much you need it to be dynamic. Gulp is pure JS, so you can very easily write your own functions. you can use the gulp-rename plugin to rename part or all of the file name before saving.

Here's a rough idea to get you started:

var rename = require('gulp-rename'),
    path = require('path'),
    glob = require('glob'); // npm i --save-dev glob    

var components = glob.sync('components/*').map(function(componentDir) {
        return path.basename(componentDir);
    });

components.forEach(function(name) {
    gulp.task(name+'-style', function() {
        return gulp.src('components/'+name+'/styles.scss')
            .pipe(sass()) // etc
            .pipe(rename(name + '.css'))
            .pipe(gulp.dest('public/assets/css'))
    });

    gulp.task(name+'-js', function() {
        // similar idea for JS files
    });

    gulp.task(name+'-build', [name+'-style', name+'-js']);
});

// build all components
gulp.task('build-components', components.map(function(name){ return name+'-build'; }));

Now you'll have tasks named component_1-build, component_1-style, component_1-js, etc, for each component.

You also have a task that can build all components.

Old question i know, but i been playing around in this area today, and hence came across this question. This may help someone if you are wanting to do something dynamic with the destination path.

You can supply a function as the argument to gulp.dest and hence do some pre processing on the destination path like this:

.pipe(gulp.dest(function (file) {
    // file is the current file in the stream
    // do something here               
    return pathString;
}));

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