I have this directory structure for a TypeScript app.
app/
--/components/
--/component1/
--component1.ts
--/scripts/
--entry.ts
.tmp/
gulpfile:
const gulp = require('gulp');
const ts= require('gulp-typescript');
If I do this:
gulp.task('typescript', function () {
return gulp.src('app/**/*.ts')
.pipe(ts())
.pipe(gulp.dest('.tmp/scripts'))
});
All typescript files generate their respective JavaScript files into the .tmp directory, with the same file structure as is in the main app:
.tmp/
--/components/
--/component1/
--component1.js
--/scripts/
--entry.js
But I want to create one output file. If I do this:
gulp.task('typescript', function () {
return gulp.src('app/**/*.ts')
.pipe(ts({
out: 'app.js'
}))
.pipe(gulp.dest('.tmp/scripts'));
});
.tmp/scripts/app.js is generated, but it only includes code from "entry.ts" - nothing from "component1.ts"
What am I doing wrong? new to gulp.
What am I doing wrong?
Usage of out
is not recommended : https://basarat.gitbooks.io/typescript/content/docs/tips/outFile.html