Gulp - copy and rename a file

JavascriptGulpGulp Watch

Javascript Problem Overview


I'm extremely new to Gulp. I'm basically trying to watch for a modified JavaScript file, and then make a new copy of it with a new name. (eventually there'll be some processing on it, but Rome wasn't built in a day).

My (naive) attempt is this:

gulp.task('default', function() {

    return gulp.watch('../**/**.js', function(obj){
        gulp.src(obj.path)
            .pipe(gulp.dest('foobar.js'));
    });

});

This takes the modified file and successfully copies it into a folder now called foobar.js. Is there anything simple I can replace gulp.dest('foobar.js') with that will simply copy and rename the src file in place?


EDIT

By copy in place, I mean I want to take the modified file, and make a copy of it right where it currently is with a new name. The equivalent of clicking the file (in windows) and hitting control-c control-v, then renaming the resulting file.

Javascript Solutions


Solution 1 - Javascript

I'm not 100% certain what you mean by

> copy and rename ... in place

But, based on your current code, if you simply wish to:

  1. Watch all .js files in the parent directory and
  2. Copy them to the cwd (current working directory) and
  3. Name all copies, regardless of source file, the same thing

Then you could use gulp-rename to do just that:

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

gulp.task('default', function() {
  return gulp.watch('../**/**.js', function(obj) {
    gulp.src(obj.path)
      .pipe(rename('newFileName.js'))
      .pipe(gulp.dest('.'));
  });
});

In this case, the output filename is newFileName.js

In order to use the module, you'll need to install the gulp-rename package with npm (ie: npm install gulp-rename).

More examples are available on the package details page on npm @ <https://www.npmjs.com/package/gulp-rename#usage>

Solution 2 - Javascript

It wasn't pretty getting there, but in the end it appears this is what I want (with some ES6 transpiling in the middle).

The key appears to be the options object with a base property in the call to src. That seems to be what's needed to maintain the path of the current file in the call to dest.

var gulp = require('gulp'),
    rename = require('gulp-rename'),
    babel = require('gulp-babel');

gulp.task('default', function() {
    return gulp.watch('../**/$**.js', function(obj){
        if (obj.type === 'changed') {
            gulp.src(obj.path, { base: './' })
                .pipe(babel())
                .pipe(rename(function (path) {
                    path.basename = path.basename.replace('$', '');
                }))
                .pipe(gulp.dest(''));
        }
    });
});

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAdam RackisView Question on Stackoverflow
Solution 1 - Javascriptknksmith57View Answer on Stackoverflow
Solution 2 - JavascriptAdam RackisView Answer on Stackoverflow