Gulp: Copy all files but exclude one extension

Question

I'm working on a Typescript project, and I am trying to implement Gulp. In my src/ folder, I have files with different extensions.

I've configured Gulp to transpile every *.ts file to Javascript using Babel, and to output the type-definition files using the Typescript compiler. I would like all other files to be copied to the dist/ folder.

const {
  dest,
  src
} = require("gulp");

function cp() {
  return src("src/**/*.*", "!(src/**/*.ts)")
    .pipe(dest("./dist"));
};

exports.cp = cp;

I would like the cp function to copy every file except those with a .ts extension to the dist/ folder.

The above code works if I define the extensions that I want to copy and avoid using the wildcard extension. I've found many examples online, but they are usually not using a wildcard extension. I'm not sure if the problem comes from my negation or from the usage of a wildcard extension. Is there any way to do this?

Many thanks!

Answer 1

Try

"!src/**/*.ts"

Remove the enclosing parentheses that you had "!(src/**/*.ts)" .

Answer 2

Found it!

The solution was written black on white in the official documentation of the src() method. Here is the fix:

const {
  dest,
  src
} = require("gulp");

function cp() {
  return src(["src/**/*.*", "!src/**/*.ts"])
    .pipe(dest("./dist"));
};

exports.cp = cp;