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Console Secrets: 5 Chrome DevTools Console APIs you probably aren't using!

  Рет қаралды 4,559

Jarrod Overson

Jarrod Overson

Күн бұрын

The chrome devtools are the first stop to debug JavaScript web applications and they include a ridiculous number of features, far more than anyone can keep up with at any point. Even if you are up-to-date, Google developers add new debugging and troubleshooting features almost every version so it's important to brush up on them regularly.
This video goes over new and old-but-unknown console like $0-$4, $_, monitorEvents()/unmonitorEvents(), getEventListeners(), monitor()/unmonitor(), and debug()/undebug(). Plus there's an undocumented feature of debug() that you'll definitely want to know about.
Check out full documentation here:
developers.goo...

Пікірлер: 12
@domaincontroller
@domaincontroller 3 жыл бұрын
00:48 sign-in form 01:30 $0, $1, $[...] variables 03:26 monitorEvents() 05:09 getEventListerners($0), unmonitorEvents() 06:28 monitor() 07:19 debug()
@JackBrunsion
@JackBrunsion 4 жыл бұрын
Absolute gold-mine!
@Surroundize
@Surroundize 4 жыл бұрын
Very helpful API. Thanks for the info. 👍🏻
@jsoverson
@jsoverson 5 жыл бұрын
If you find any other undocumented tips with console methods please post them!
@logkdogk
@logkdogk 4 жыл бұрын
Awesome video! Thanks for the headsup! 🤘🤘🤘
@mirhosseinmousavi7378
@mirhosseinmousavi7378 Жыл бұрын
Really cool stuff! thanks
@user-cu8sv7lg7f
@user-cu8sv7lg7f 8 ай бұрын
Very helpful API!!!😀😀😀😀
@MrFedericoraggi
@MrFedericoraggi 2 жыл бұрын
Super useful. Thank you very much
@thetrends5670
@thetrends5670 3 жыл бұрын
Who draw those drawings in your background/wall?
@osherezra8460
@osherezra8460 4 жыл бұрын
Wow Thank you so much was looking for an answer to the $0 for ages. just one question why is that every time you run the command it return undefined? cheers
@mishasawangwan6652
@mishasawangwan6652 3 жыл бұрын
i’m going off on a guess here but i think it’s because if not explicitly specified every function/expression/whatever in javascript has an implicit return value of ‘undefined’ therefore ..well, i think the rest is obvious..
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