Can Gulp Overwrite All Src Files?


Answer :

I can think of two solutions:

  1. Add an option for base to your gulp.src like so:

    gulp.src([...files...], {base: './'}).pipe(...)...

    This will tell gulp to preserve the entire relative path. Then pass './' into gulp.dest() to overwrite the original files. (Note: this is untested, you should make sure you have a backup in case it doesn't work.)

  2. Use functions. Gulp's just JavaScript, so you can do this:

    [...files...].forEach(function(file) {     var path = require('path');     gulp.src(file).pipe(rename(...)).pipe(gulp.dest(path.dirname(file))); }

    If you need to run these asynchronously, the first will be much easier, as you'll need to use something like event-stream.merge and map the streams into an array. It would look like

    var es = require('event-stream');  ...  var streams = [...files...].map(function(file) {         // the same function from above, with a return         return gulp.src(file) ...     }; return es.merge.apply(es, streams);

Tell gulp to write to the base directory of the file in question, just like so:

    .pipe(         gulp.dest(function(data){              console.log("Writing to directory: " + data.base);             return data.base;         })     )

(The data argument is a vinyl file object)

The advantage of this approach is that if your have files from multiple sources each nested at different levels of the file structure, this approach allows you to overwrite each file correctly. (As apposed to set one base directory in the upstream of your pipe chain)


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