How to set gulp.dest() in same directory as pipe inputs? - gulp

Question

I need all the found images in each of the directories to be optimized and recorded into them without setting the path to the each folder separately. I don't understand how to make that.
var gulp = require('gulp');
var imageminJpegtran = require('imagemin-jpegtran');
gulp.task('optimizeJpg', function () {
return gulp.src('./images/**/**/*.jpg')
.pipe(imageminJpegtran({ progressive: true })())
.pipe(gulp.dest('./'));
});

Answer 1

Here are two answers.
First: It is longer, less flexible and needs additional modules, but it works 20% faster and gives you logs for every folder.
var merge = require('merge-stream');
var folders =
[
"./pictures/news/",
"./pictures/product/original/",
"./pictures/product/big/",
"./pictures/product/middle/",
"./pictures/product/xsmall/",
...
];
gulp.task('optimizeImgs', function () {
var tasks = folders.map(function (element) {
return gulp.src(element + '*')
.pipe(sometingToDo())
.pipe(gulp.dest(element));
});
return merge(tasks);
});
Second solution: It's flexible and elegant, but slower. I prefer it.
return gulp.src('./pictures/**/*')
.pipe(somethingToDo())
.pipe(gulp.dest(function (file) {
return file.base;
}));

Answer 2

Here you go:
gulp.task('optimizeJpg', function () {
return gulp.src('./images/**/**/*.jpg')
.pipe(imageminJpegtran({ progressive: true })())
.pipe(gulp.dest('./images/'));
});
Gulp takes everything that's a wildcard or a globstar into its virtual file name. So all the parts you know you want to select (like ./images/) have to be in the destination directory.

Answer 3

You can use the base parameter:
gulp.task('uglify', function () {
return gulp.src('./dist/**/*.js', { base: "." })
.pipe(uglify())
.pipe(gulp.dest('./'));
});