Q:

CSS minify and rename with gulp

I've a variable like
var files = {
    'foo.css': 'foo.min.css',
    'bar.css': 'bar.min.css',
};

What I want the gulp to do for me is to minify the files and then rename for me.
But the tasks is currently written as (for one file)
gulp.task('minify', function () {
    gulp.src('foo.css')
        .pipe(minify({keepBreaks: true}))
        .pipe(concat('foo.min.css'))
        .pipe(gulp.dest('./'))
});

How to rewrite so it work with my variable files defined above?

A:

You should be able to select any files you need for your src with a Glob rather than defining them in an object, which should simplify your task. Also, if you want the css files minified into separate files you shouldn't need to concat them.
var gulp = require('gulp');
var minify = require('gulp-minify-css');
var rename = require('gulp-rename');

gulp.task('minify', function () {
    gulp.src('./*.css')
        .pipe(minify({keepBreaks: true}))
        .pipe(rename({
            suffix: '.min'
        }))
        .pipe(gulp.dest('./'))
    ;
});

gulp.task('default', ['minify'], function() {

});

A:

I tried the earlier answers, but I got a never ending loop because I wasn't ignoring the files that were already minified. 
First use this code which is similar to other answers:
//setup minify task
var cssMinifyLocation = ['css/build/*.css', '!css/build/*.min.css'];
gulp.task('minify-css', function() {
  return gulp.src(cssMinifyLocation)
    .pipe(minifyCss({compatibility: 'ie8', keepBreaks:false}))
    .pipe(rename({ suffix: '.min' }))
    .pipe(gulp.dest(stylesDestination));
});

Notice the '!css/build/*.min.css' in the src (i.e. var cssMinifyLocation)
//Watch task
gulp.task('default',function() {
    gulp.watch(stylesLocation,['styles']);
    gulp.watch(cssMinifyLocation,['minify-css']);
});

You have to ignore minified files in both the watch and the task. 

