How to wait for all the files, manipulate them, run them back one by one?

Question

gulp.js

gulp.scr(["js/*.html"])
.pipe(myModule("template.html"))
.pipe(gulp.dest("public"))

How to wait for all the files in myModule to change them and start again one by one as gulp.src
(this is the old assembler that moves to gulp, each file stores data for the heat, therefore, in order to form them normally, you need all the files at the same time)

Answer 1

create 2 tasks so that one is a dependency of the other.

gulp.task('taskName', [taskWhichShouldDoneBefore], function() {
 });