Use filename from earlier gulp.src in gulp.rename() call

Question

I have a zip-file in a folder, I want to extract one of its (zipped) files and give it a filename pattern source file basename + .xml in folder ./sourcefiles_unpacked/

./sourcefiles/test.zip

=>

./sourcefiles/test.zip
./sourcefiles_unpacked/test.xml

Unzipping & filtering works nicely with gulp-unzip, however I'm not sure how to access the filename from the gulp.src call.

gulp.task('unzip-filtered-rename', function() {
  return gulp.src(paths.unzip_sourcefiles)
    // .pipe(debug())
    .pipe(plumber({
      errorHandler: notify.onError('unzip-filtered-rename error: <%= error.message %>')
    }))
    .pipe(changed(paths.excel_targetdir_local_glob, {
      extension: '.xml'
    }))
    .pipe(unzip({filter : function(entry){return minimatch(entry.path, "contents.xml")}}))
    .pipe(gulp.rename(function(path){       

        // ? What do I put here to rename each target file to 
        // ?   its originating zip file's basename?


    })) //   "==> test.xml",
    .pipe(gulp.dest(paths.sourcefiles_unpacked)) //   sourcefiles_unpacked: "./sourcefiles_unpacked/"
});

As soon as gulp.rename() is called the chunk has been renamed to its name as in the zipfile.

• How do I access or store the filepath of an earlier pipe for use in the rename function call?
• Is the gulp.dest properly configured if paths contains the glob "./sourcefiles_unpacked/"?

Answer 1

Try this:

const glob = require("glob");
const zipFiles = glob.sync('sourcefiles/*.zip');

gulp.task('unzip-filtered-rename', function (done) {

  zipFiles.forEach(function (zipFile) {

    const zipFileBase = path.parse(zipFile).name;

    return gulp.src(zipFile)
        // .pipe(debug())
        //   .pipe(plumber({
        //     errorHandler: notify.onError('unzip-filtered-rename error: <%= error.message %>')
        //   }))
        //   .pipe(changed(paths.excel_targetdir_local_glob, {
        //     extension: '.xml'
        //   }))
        //   .pipe(unzip({filter : function(entry){return minimatch(entry.path, "contents.xml")}}))
        .pipe(rename({
            basename: zipFileBase,
        }))
        .pipe(gulp.dest("./sourcefiles_unpacked")) //   sourcefiles_unpacked: "./sourcefiles_unpacked/"
    });
    done();
});

I commented out the other stuff you do just for testing purposes. Running each file through a forEach or map allows you to set a variable at the beginning outside of the stream that will be available in the following stream.

Also see How to unzip multiple files in the same folder with Gulp for more discussion on setting variables to be used in a stream.