Feature Request: Specify s3 Path

[mrprkr]

Is is possible to add an option to specify the upload path? AWS API suggests prepending the file name with /folder/to/upload/to/. Can't figure out how to implement this here.

[clineamb]

Hello! Checkout the keyTransform function:

keyTransform (nameTransform)

Type: function

Use this to transform your file names before they're uploaded to your S3 bucket.
(Previously known as name_transform).

    gulp.task("upload_transform", function() {
        return gulp.src("./dir/to/upload/**")
            .pipe(s3({
                Bucket: 'example-bucket',
                ACL: 'public-read',
                keyTransform: function(relative_filename) {
                    var new_name = changeFileName(relative_filename);
                    // or do whatever you want
                    return new_name;
                }
            }))
        ;
    });

So, if you want to prepend the folder name...

    gulp.task("upload_transform", function() {
        return gulp.src("./dir/to/upload/**")
            .pipe(s3({
                Bucket: 'example-bucket',
                ACL: 'public-read',
                keyTransform: function(relative_filename) {
                    var new_name = "folder/to/upload/to/" + relative_filename;
                    // or you can use `path.join("folder", "to", "upload", relative_filename)`
                    return new_name;
                }
            }))
        ;
    });

[mrprkr]

Great, thanks for this!

Sent from my iPhone

On 5 Jan 2016, at 8:37 AM, Caroline A. [email protected] wrote:

Hello! Checkout the keyTransform function:

keyTransform (nameTransform)

Type: function

Use this to transform your file names before they're uploaded to your S3 bucket.
(Previously known as name_transform).

gulp.task("upload_transform", function() {
    gulp.src("./dir/to/upload/**")
        .pipe(s3({
            Bucket: 'example-bucket',
            ACL: 'public-read',
            keyTransform: function(relative_filename) {
                var new_name = changeFileName(relative_filename);
                // or do whatever you want
                return new_name;
            }
        }))
    ;
});

So, if you want to prepend the folder name...

gulp.task("upload_transform", function() {
    gulp.src("./dir/to/upload/**")
        .pipe(s3({
            Bucket: 'example-bucket',
            ACL: 'public-read',
            keyTransform: function(relative_filename) {
                var new_name = "folder/to/upload/to/" + relative_filename;
                // or you can use `path.join("folder", "to", "upload", relative_filename)`
                return new_name;
            }
        }))
    ;
});

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[vladaman]

This method is great but it does not allow me to specify prefix for example. Is there any way to keep the folder structure and prefix all uploads with a path?

 gulp.src(["index.html",'./assets/**'])
        .pipe(s3({
            Bucket: 'mybucket',
            ACL: 'public-read',
            keyTransform: function(relative_filename) {
                var new_name = "production/" + relative_filename; // causes issue, since all keys will be prefixed but original path is not used
                return new_name;
            }
        }, 

[kgraf80]

This is my solution (workaround)
in /node_modules/gulp-s3-upload/index.js change file.relative to: file.path

if(keyTransform) {

    //  Allow the transform function to take the
    //  complete path in case the user wants to change
    //  the path of the file, too.

    // >>>>>>> change here: file.relative to: file.path
    keyname = keyTransform(file.path);

} else {
    // ...Otherwise keep it exactly parallel.

    keyparts = helper.parsePath(file.relative);
    keyname  = helper.buildName(keyparts.dirname, keyparts.basename + keyparts.extname);
}

and after that:

const targetAssets = 'aws-assets/';

 gulp.src(["index.html",'./assets/**'])
        .pipe(s3({
            Bucket: 'mybucket',
            ACL: 'public-read',
            keyTransform: function(filePath) {
                let newFile = targetAssets + filePath.split("www/assets/")[1]
                return newFile;
            }
        },

for example:
for filePath: /home/user/webserwer/www/assets/images/test/image.jpg
file will be saved in: mybucket/aws-assets/images/test/image.jpg