Only show console.log if debug is set to true

[KirianCaumes]

Hello,

As I'd rather handle console.log myself, what do you think about hiding them in the index.js if debug is not set to true:

//...
async fetch() {
    this.session.debugOn = this.debug;
    await this.session.fetch();
    if (this.session.status === 'OK') {
        this.video = new Video(this.session, this.options);
        this.video.debugOn = this.debug;
        await this.video.fetch();
        if (this.video.status === 'OK') {
            if (this.debug)
                console.log(ut.jsp(this.video.info(exclusions)));
            await this.video.download();
            if (this.video.downloaded) {
                if (this.debug) {
                    console.log('\n\nDone!');
                    console.log(`Downloaded: ${this.video.fn}`);
                }
            } else {
                throw new Error(`Video status: ${this.video.status} (${this.video.reason})`);
            }
        } else {
            throw new Error(`Video status: ${this.video.status} (${this.video.reason})`);
        }
    } else {
        throw new Error(`Session status: ${this.session.status} (${this.session.reason})`);
    }
}
//...

As you can see, I've also put some throw new Error('...'), which allow my script to do the error handling.

Thank you, awesome lib!

[gatecrasher777]

Thanks for the support. What I would suggest you do in your code instead of editing index.js is create your own main.js, extend the Download class and overwrite the fetch() method. Something like this.

const ytcog = require('ytcog');

class MyDL extends ytcog.Download {

  async fetch() {
      this.session.debugOn = this.debug;
      await this.session.fetch();
      if (this.session.status === 'OK') {
          this.video = new ytcog.Video(this.session, this.options);
          this.video.debugOn = this.debug;
          await this.video.fetch();
          if (this.video.status === 'OK') {
              this.video.debug(this.video.info());
              await this.video.download();
              if (this.video.downloaded) {
                  this.video.debug('\n\nDone!');
                  this.video.debug(`Downloaded: ${this.video.fn}`);
              } else {
                  throw new Error(`Video status: ${this.video.status} (${this.video.reason})`);
              }
          } else {
              throw new Error(`Video status: ${this.video.status} (${this.video.reason})`);
          }
      } else {
          throw new Error(`Session status: ${this.session.status} (${this.session.reason})`);
      }
   }

}

let download = new MyDL(options, cookie, userAgent, proxy, debug);
await download.fetch();

[KirianCaumes]

That's perfect for me that way, thank you!

I found something else : some video that I downloaded have don't have sound. Can we discuss that here, or would you prefer that I open a new issue?

[gatecrasher777]

You can close this and open an new issue. If you can give me an example that I can replicate then I can fix asap.