Excluding a file change in gulp but still keeping the file

Question

The following is my simple gulp task:

• gulp.src is getting script.js as input
• minifies script.js and saves script.min.js
• removes console.log from both script.js and script.min.js
• updates script.js file and saves script.min.js

I want to simply remove console.log from a minified file (or script.min.js) not from script.js, and I am failing. I am looking for a way to run .pipe(gulpRemoveLogging... only on a minified file (i.e. script.min.js) and do not modify anything from script.js

var gulp = require("gulp"),
minify = require("gulp-minify"),
gulpRemoveLogging = require("gulp-remove-logging");

var _dist_path = "./dist/"; // relative path

// minify JS files
gulp.task("minify-js", function() {
    return gulp.src(path.join(_dist_path, "/*.js"))
        .pipe(minify({
            ext: {
                src: ".js",
                min: ".min.js"
            },
            ignoreFiles: ["-min.js"]
        }))
        .pipe(gulpRemoveLogging({
            namespace: ["console", "window.console"]    // remove these namespaces. e.g. console.log("..."), window.console.log("...")
        }))
        .pipe(gulp.dest(_dist_path));
});

Answer 1

Solved the problem:

Added .pipe(gulpIgnore("smartsitescripts.js")) between .pipe(minify({ ... and .pipe(gulpRemoveLogging ...

var gulpIgnore = require("gulp-ignore");

...
    .pipe(minify({
        ext: {
            src: ".js",
            min: ".min.js"
        },
        ignoreFiles: ["-min.js"]
    }))
    .pipe(gulpIgnore("script.js"))  // <---- this line solved the problem
    .pipe(gulpRemoveLogging({
        namespace: ["console", "window.console"]    // remove these namespaces. e.g. console.log("..."), window.console.log("...")
    }))
...

That line of code will exclude script.js from stream hence stream will contain only script.min.js.