Apply conditions in gulp.dest - gulp

Question

I have a gulp task which gets the html file, finds the scripts inside it, minifies them, applies revisions and outputs everything in the assets folder:
templates/index.html
<!-- build:js js/app.js -->
<script src="lib/js/a.js"></script>
<script src="lib/js/b.js"></script>
<!--endbuild -->
gulpfile.js
var useref = require('gulp-useref'),
filter = require('gulp-filter'),
uglify = require('gulp-uglify'),
rev = require('gulp-rev'),
revReplace = require('gulp-rev-replace');
gulp.task('build',function() {
var assets = useref.assets({searchPath: './'}),
jsapp = filter('**/app.js'),
return gulp
.src(gulp.src('templates/index_src.html'))
// collect all assets from source file by the means of useref
.pipe(assets)
//build js/app.js
.pipe(jsapp)
.pipe(uglify())
.pipe(jsapp.restore())
// Take inventory of the file names for future rev numbers
.pipe(rev())
// Apply the concat and file replacement with useref
.pipe(assets.restore())
.pipe(useref())
// Replace the file names in the html with rev numbers
.pipe(revReplace())
// output files
.pipe(gulp.dest('./'));
});
This works fine but outputs everything (js/app.js and index.html) to the root directory (./);
Is there are any way to apply a condition inside the gulp.dest and output js/app.js to the root folder of the project but output index.html to a different location (e.g. ./templates_cache)?

Answer 1

Also add a filter for your html files, to be able to output only those files to a different folder:
var htmlFilter = filter('**/*.html')
Change your output to:
// output for js files
.pipe(jsapp)
.pipe(gulp.dest('./'))
.pipe(jsapp.restore())
// output for html files
.pipe(htmlFilter)
.pipe(gulp.dest('./templates_cache'));