Hi, I'm trying to replace a script include with an inline script of it's contents.
gulp.task('replace-inline', ['minify'], function() {
return gulp.src(paths.distRoot+'index.web.html')
.pipe(replace('<script src="js/async.min.js"></script>', '<script><%= gulp.file.read(paths.distRoot+\'/js/async.min.js\') %></script>'))
.pipe(gulp.dest(paths.distRoot+'index.web.html'));
});
Not too sure about gulps inner working, idea how I might be able to do it?
Hi, I'm trying to replace a script include with an inline script of it's contents.
Not too sure about gulps inner working, idea how I might be able to do it?