Thanks bro for your tutorials. They have been the best of all that I have tried to attach myself to. Keep us updated. Lots of appreciation to Giraffe Academy
Fuck I'm an idiot. Mike at 3:57: "That's probably because David Wallace makes 250 grand a year." Me: David Wallace... Where do I know that from? Oh! The Office! Coincidence i guess. *Looks at the other names* Nope they're all there, I'm an idiot.
Hi, I'm following your series of tutorials and I have a question for you. I'm in this line of code SELECT SUM(total_sales), emp_id FROM works_with GROUP BY emp_id; the question is: how do i get the output of the name of the employee instead of their emp_id?. We are working on the works_with table but the name is in the employee table.
OK, nevermind, I got to the JOINS video tutorial (17) and solved it with this ------------------------------------------------------------------------------------------------------------- SELECT employee.emp_id, employee.first_name, SUM(total_sales) FROM employee JOIN works_with ON employee.emp_id = works_with.emp_id GROUP BY employee.emp_id;
SELECT e.first_name, e.last_name FROM employee AS e WHERE emp_id = (SELECT emp_id FROM works_with); This would connect emp_id with works with and employee and return all employees who are included in the works_with table. You can do a nested query like this or a JOIN statement I believe for this.
Hi, I need that the values for a ColumnID, which is a Primary Key, to start at 100 and increment by 5. AUTO_Increment doesn't work in my case because it increases only by 1. How can I do that in MySQL? Hope you could help me. Thanks.
Hi Zirely. Follow this tutorial www.tutorialspoint.com/mysql-what-is-auto-increment-5-in-a-create-table-query#:~:text=The%20AUTO_INCREMENT%3D5%20in%20a,starts%20from%201%20by%20default. in case you want to thanks or ask further, subscribe to my channel. I will help you.
The code of the last video has the following part: -- Find all employee's id's and names who were born after 1969 SELECT emp_id, first_name, last_name FROM employee WHERE birth_day >= 1970-01-01; should it be WHERE birth_day >= 1970-01-01; or WHERE birth_day >= '1970-01-01'; (add a single quotes to 1970-01-01) ?
Guest wants to display Resort Id, Name and Comments of the resort based on it's star rating. Comments of the resort displays as follows If rating is between 5.0 to 4.5 then display the comments as 'Excellent Resort' If rating is between 4.4 to 4.0 then display the comments as 'Very Good Resort' else display 'Good Resort' . Give alias name to this result as Comments. Sort the result based on resort id pls help me with this question and also help me with some tricks.
@@jaychang5631 I see what it is now. Some of the employees are listed twice in the Works With table (for multiple sales). As for your comment about groups, he did use "GROUP BY emp_id"... If you pause at 7:22, it's there.
Thanks bro for your tutorials. They have been the best of all that I have tried to attach myself to. Keep us updated.
Lots of appreciation to Giraffe Academy
Thanks for sharing..very helpful! Love the "Office" employees! :)
This is the coolest one.
Mike Dane + SQL =
This is Great. The best place to start SQL. Thanks Man!
Unfortunately he titled it 14....so I am asking where is 1-13
ruclips.net/video/2KbZ12RK_OI/видео.html I also tried explaining this topic. Please let me know any feedback.
Fuck I'm an idiot.
Mike at 3:57: "That's probably because David Wallace makes 250 grand a year."
Me: David Wallace... Where do I know that from? Oh! The Office! Coincidence i guess. *Looks at the other names* Nope they're all there, I'm an idiot.
his code that was originally copied was birth_day not birth_date - make sure you know that
Mike is testing us to see if we are paying attention ;)
@@zen123w yea, I was like why is mine not working??? I know it's in there right. Took me a min and I went back through the data LOL
you're the best
you didn't use joins in the last example, how?
this helps a lot!!
ruclips.net/video/2KbZ12RK_OI/видео.html I also tried explaining this topic. Please let me know any feedback.
♥️♥️🙏
Hi, I'm following your series of tutorials and I have a question for you. I'm in this line of code
SELECT SUM(total_sales), emp_id
FROM works_with GROUP BY emp_id;
the question is: how do i get the output of the name of the employee instead of their emp_id?. We are working on the works_with table but the name is in the employee table.
OK, nevermind, I got to the JOINS video tutorial (17) and solved it with this
-------------------------------------------------------------------------------------------------------------
SELECT employee.emp_id, employee.first_name, SUM(total_sales)
FROM employee
JOIN works_with
ON employee.emp_id = works_with.emp_id
GROUP BY employee.emp_id;
Such a great tutorial!
How to connect emp_id with their name from a different table?
SELECT e.first_name, e.last_name FROM employee AS e WHERE emp_id = (SELECT emp_id FROM works_with);
This would connect emp_id with works with and employee and return all employees who are included in the works_with table. You can do a nested query like this or a JOIN statement I believe for this.
ruclips.net/video/2KbZ12RK_OI/видео.html I also tried explaining this topic. Please let me know any feedback.
Awesome
It must be good to work at Dunder Mifflin 😆
>= '1971-01-01' or > '1970-12-31'
How to find which employee sales high and which client buy more??
Could you make video on lowercase letters and uppercase
I need more clarification on how to relate tables using SQL
Gonna have to go through this again. I understand somewhat, but I don't understand why it works. :(
Hi, I need that the values for a ColumnID, which is a Primary Key, to start at 100 and increment by 5. AUTO_Increment doesn't work in my case because it increases only by 1. How can I do that in MySQL? Hope you could help me. Thanks.
Hi Zirely. Follow this tutorial www.tutorialspoint.com/mysql-what-is-auto-increment-5-in-a-create-table-query#:~:text=The%20AUTO_INCREMENT%3D5%20in%20a,starts%20from%201%20by%20default.
in case you want to thanks or ask further, subscribe to my channel. I will help you.
The code of the last video has the following part:
-- Find all employee's id's and names who were born after 1969
SELECT emp_id, first_name, last_name
FROM employee
WHERE birth_day >= 1970-01-01;
should it be WHERE birth_day >= 1970-01-01; or WHERE birth_day >= '1970-01-01'; (add a single quotes to 1970-01-01) ?
obviously : >= '1971-01-01' or > '1970-12-31' and yes you need quotes since DATE FORMAT is rather String. Or else that would be a numeric sequence.
ruclips.net/video/2KbZ12RK_OI/видео.html I also tried explaining this topic. Please let me know any feedback.
Guest wants to display Resort Id, Name and Comments of the resort based on it's star rating.
Comments of the resort displays as follows
If rating is between 5.0 to 4.5 then display the comments as 'Excellent Resort'
If rating is between 4.4 to 4.0 then display the comments as 'Very Good Resort'
else display 'Good Resort' . Give alias name to this result as Comments.
Sort the result based on resort id
pls help me with this question and also help me with some tricks.
mysql
m=mike
y=youth
s=structure
q=query
l=languag...
Hey Mohit, let's connect it's very easy
7:22 Why did only only four employees show up in the table if there are nine employees from the works_with table?
maybe other employees are not salesman. BTW, seems the video use wrong group_by, should be "GROUP BY
emp_id;" right?
@@jaychang5631 I see what it is now. Some of the employees are listed twice in the Works With table (for multiple sales).
As for your comment about groups, he did use "GROUP BY emp_id"... If you pause at 7:22, it's there.
On female employees born after 1971 I don't h=think we got the right answer there.
none of this is a function!!!
SUM(), COUNT(), AVG() are.