Question:
Here is an example
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('./'));
});
I need all found images in this directory to be optimized and written to it. But I don't quite understand how it could be done.
Answer:
gulp.dest
supports using a function as the first argument. This function takes as argument a file as a vinyl-объект
. The string it returns is used as the save directory.
Try like this :
gulp.dest(function(file){
return file.base;
})