Anyone noticed? "If your answer is engaging and lengthy the interviewer forgets the other one question" 😂😂😂 Tushar plays the game to not answer that question which i think he doesn't know or is not interested in.
@@eng.procodrr🤣 great video, different questions asked. Need videos on debugging issues and advanced concepts questions in react.js for interview. experienced
Sir i want to give you the interview , I'm from Pakistan but my pc is very laggy and it doesn't have the camera. But I'm only available at night 10:30 because there is alot of load shedding in our Area
Sir I would also like to give an interview. I am watching this series for a approx and I am also preparing for frontend interview. So, for reviewing my skills I want to give an interview please sir.
Candidate interviewer ki sunn hi nhi tha rha tha, ye galat h thda, theek h aapko aata h but agar samne wala kuch bol rha h to you have to listen carefully
Jab aap yeh code browser mein run karte ho, toh name ek global variable ban jata hai, jo ki browser ki window object ka ek property hota hai. Browser mein, by default, yeh empty string hota hai. Lekin jab aap var name = "Rajat"; statement ko execute karte ho, toh aap basically ek naya variable define kar rahe ho, jiska naam name hai aur uski value "Rajat" hai. Isi liye browser mein console.log(name); "Rajat" print karta hai. Lekin, jab aap yehi code terminal ke console mein run karte ho, toh situation thodi alag hoti hai. Terminal ke console mein name variable ka koi predefined value nahi hoti, aur jab aap console.log(name); ke upar aate ho, tab name variable toh hoisted ho jaata hai, lekin uski koi value nahi hoti. Isliye yeh undefined print karta hai.
Jab aap yeh code browser mein run karte ho, toh name ek global variable ban jata hai, jo ki browser ki window object ka ek property hota hai. Browser mein, by default, yeh empty string hota hai. Lekin jab aap var name = "Rajat"; statement ko execute karte ho, toh aap basically ek naya variable define kar rahe ho, jiska naam name hai aur uski value "Rajat" hai. Isi liye browser mein console.log(name); "Rajat" print karta hai. Lekin, jab aap yehi code terminal ke console mein run karte ho, toh situation thodi alag hoti hai. Terminal ke console mein name variable ka koi predefined value nahi hoti, aur jab aap console.log(name); ke upar aate ho, tab name variable toh hoisted ho jaata hai, lekin uski koi value nahi hoti. Isliye yeh undefined print karta hai.
Tushar looks confident and has a good knowledge about how to give the Interview❤️😍
Anyone noticed? "If your answer is engaging and lengthy the interviewer forgets the other one question"
😂😂😂
Tushar plays the game to not answer that question which i think he doesn't know or is not interested in.
Every time when i watch the interview i learn new things........... thank you so much
Love U Sir ❤
Thank for all. Plz.. Not Stop Your Journey With Us
Thankyou for the valuable session !!!😃
It was really nice interviewing you Tushar 😃
Jhon was epic 😂
Yes, that was really epic 😂
@@eng.procodrr🤣 great video, different questions asked. Need videos on debugging issues and advanced concepts questions in react.js for interview. experienced
Sir i want to give you the interview , I'm from Pakistan but my pc is very laggy and it doesn't have the camera. But I'm only available at night 10:30 because there is alot of load shedding in our Area
I think he have proven experience.
Congratulations sir ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
Awsome interview
The answer john will be historical.......i have never seen such🤣🤞🏻
True 😂
Do freelance series brother, how to get clients on freelancing.
sara kuch chodoo sir ka flow dekho english ka😍😍😍
Guy has more experience of using js libs than writing (core-fundamental) js :p
sir please mujhe yeh batao ki mujhe html css js react aata hai to mei kya frontend dev ke liye apply karu ya kuch aur sikhna padega please reply me
You can start applying.
Build portfolio
Try to learn basic animations
Like gsap.. Or use css,js
@@ReelFuseBox_YT mujhe samajh nahi aaya aap kya khna chate ho aap animations ki baat kar rahe ho kya
Sir angular pay bhi eik series bana dijiye plz
How i can give mock interview on your channel sir , (fresher )
Sir I would also like to give an interview. I am watching this series for a approx and I am also preparing for frontend interview. So, for reviewing my skills I want to give an interview please sir.
Sir mujhe java,html,css, javascript,mysql ata hai to kya frontend ke kiye yeh kafi hai
React seekh lo
Candidate interviewer ki sunn hi nhi tha rha tha, ye galat h thda, theek h aapko aata h but agar samne wala kuch bol rha h to you have to listen carefully
19:54 don't make oversmart and laugh in front of the interviewer guy 😂😂😂 he knows everything
Sir python por video banye
"john"
😃😃
Jab aap yeh code browser mein run karte ho, toh name ek global variable ban jata hai, jo ki browser ki window object ka ek property hota hai. Browser mein, by default, yeh empty string hota hai. Lekin jab aap var name = "Rajat"; statement ko execute karte ho, toh aap basically ek naya variable define kar rahe ho, jiska naam name hai aur uski value "Rajat" hai. Isi liye browser mein console.log(name); "Rajat" print karta hai.
Lekin, jab aap yehi code terminal ke console mein run karte ho, toh situation thodi alag hoti hai. Terminal ke console mein name variable ka koi predefined value nahi hoti, aur jab aap console.log(name); ke upar aate ho, tab name variable toh hoisted ho jaata hai, lekin uski koi value nahi hoti. Isliye yeh undefined print karta hai.
ye real channel hai?
Very very basic question but i enjoyed
john🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
John
john🤣
i laugh so hard
Jab aap yeh code browser mein run karte ho, toh name ek global variable ban jata hai, jo ki browser ki window object ka ek property hota hai. Browser mein, by default, yeh empty string hota hai. Lekin jab aap var name = "Rajat"; statement ko execute karte ho, toh aap basically ek naya variable define kar rahe ho, jiska naam name hai aur uski value "Rajat" hai. Isi liye browser mein console.log(name); "Rajat" print karta hai.
Lekin, jab aap yehi code terminal ke console mein run karte ho, toh situation thodi alag hoti hai. Terminal ke console mein name variable ka koi predefined value nahi hoti, aur jab aap console.log(name); ke upar aate ho, tab name variable toh hoisted ho jaata hai, lekin uski koi value nahi hoti. Isliye yeh undefined print karta hai.