I have two tables one
Employee and mailing Subscriptions
Employee looks like this:
Name (pk) | Surname | Age
mailing Subsriptions
MailId (pk)| EmployeeName (fk)|Description | Date
I wanted to subscription number for each customer, therefore I tried the following query:
Select COUNT(c.Name)
From Employee
INNER JOIN mailingSubscriptions as m ON c.Name = m.EmployeeName;
It will give me all counts for each Employee that has an entry in the mailing subscription.
My problem is that I want to see the counts for ALL the Employees, including the ones without an entry (therefore to show 0), I tried an outer left/right join, but it will not work. Am I doing something wrong?
SELECT c.name, count(m.mailid)
FROM Employee
LEFT JOIN mailingSubscriptions as m ON c.Name = m.EmployeeName
GROUP BY c.name;
Related
I have one table customer, and one bill, and one sell.
Customer table
id-----name
Bill table
id-----customer_id
Sell table
id-----customer_id-----bill_id-----qtt-----price
A customer can have the records in the sell table with customer_id, and also have the bill record in the bill table with customer_id and this bill record has record in the sell table with bill_id.
This means a customer can have direct or indirect (in this case by passing the bill table) with sell table.
Now how to join tables that retrieve the total sell of a customer with ascending or deciding order?
Any idea?
I have tried many ways for example something like below, but none of them was working:
SELECT
sell.id AS sell_id,
customer.id,
bill.id AS bill_id,
customer.`name`,
sell.quantity*sell.price AS sell_price
FROM
customer_tb customer
JOIN bill ON bill.customer_id = customer.id
JOIN sell ON sell.customer_id = customer.id OR sell.bill_id = bill.id
NOTE: In case of bill table has a record in the sell table, the customer_id column is NULL and also same for the customer sell record the bill_id is NULL, this means in the sell table in the same entry only one of the (customer_id, bill_id) column has value.
You have the customer column in your sell tables, you don't need to join with bill table.
SELECT
customer.id,
customer.name,
SUM(sell.quantity * sell.price) AS total_amount
FROM customer_tb as customer
INNER JOIN sell
On sell.customer_id = customer.id
GROUP BY customer.id
You have to group your rows by the customer to get their sum independently
Your challenge is to look up the correct Customer.name for each Sell row.
You need to join Sell to Customer, and also join Sell to Bill to Customer.
How to do this?
FROM Sell s
LEFT JOIN Cust c1 ON s.customer_id = c1.id
LEFT JOIN Bill b on s.bill_id = b.id
LEFT JOIN Cust c2 ON b.customer_id = c2.id
That gives us:
s.id: the Sale id
s.quantity, s.price: the business details of the Sale
c1.name: the Customer name directly from the Sale's customer_id
c2.name: the Customer name indirectly through the Sale's bill_id
Because we use LEFT JOINs, either name can be NULL depending on which ids are null. And, I believe you favor using the direct name over the indirect name if you have both.
That means we use COALESCE() to choose one name from the two. Something like this
SELECT s.id, s.Quantity, s.Price,
COALESCE (c1.name, c2.name) name,
COALESCE (c1.id, c2.id) customer_id
FROM Sell s
LEFT JOIN Cust c1 ON s.customer_id = c1.id
LEFT JOIN Bill b on s.bill_id = b.id
LEFT JOIN Cust c2 ON b.customer_id = c2.id
But, if you have a chance to rework this database before you put a ton of data into it, please consider redesigning things to get rid of this ambiguity. It's going to drive you mad and cost you bigger servers if you have to make it work well with megarows.
I am working on writing a SQL query to produce a table that will look something like this:
Name |Dates Absent|Total Absences
student |10/28/2018 | 2
|10/29/2018 |
I currently have a data base which has 2 tables that are part of a larger system which contain the needed data (absences, students).
I have tried the following query
SELECT s.student_id,s.last_name,s.first_name, COUNT(s.student_id) AS 'Total Absences'
FROM `students` s, `absences` a INNER JOIN students ON students.student_id=a.student_id
Which yielded the following results:
student_id | last_name | first_name | Total Absences
1 | student | name | 12464
I want this to only use each ID once and count the times it appears. Is the problem from a relationship in the database that has many dates the one person can be absent? The ID was left in the select for now for debugging purposes, it will be removed later.
EDIT
I now have the query
SELECT s.last_name, s.first_name,a.date_absence, COUNT(s.student_id) AS 'Total Absences'
FROM `students` s, `absences` a
INNER JOIN students ON students.student_id=a.student_id
GROUP BY s.student_ID
This only displays one of the dates, how I can add all of the dates without redisplaying the students information?
You can do this with group_concat. It's not quite what you descibe, but it's close.
SELECT s.student_id,s.last_name,s.first_name, group_concat(a.date_absent) AS 'Dates Absent', COUNT(a.id) AS 'Total Absences'
FROM `students` s JOIN `absences` a ON s.student_id = a.student_id
GROUP BY s.student_id
which should yield
student_id | last_name | first_name | Dates Absent | Total Absences
1 | student | name | 10/28/2018,10/29/2018 | 2
It looks like you are almost there with the counting, but missing your GROUP BY statement
If you include aggregate functions, such as COUNT(), but leave off the GROUP BY, the whole intermediate result is taken as one group
You also seem to have a strange CROSS JOIN going on with your duplicate mention of the students table
If you want the absence dates in each row you'll have to use another aggregate function, GROUP_CONCAT()
Something along the lines of
SELECT s.student_id, /** Include as names could feasibly be duplicated */
CONCAT(s.first_name, ' ', s.last_name) name,
GROUP_CONCAT([DISTINCT] a.date) dates_absent, /** May want distinct here if more than one absence is possible per diem */
COUNT(*) total_absences
FROM students s
JOIN absences a
ON a.student_id = s.student_id
GROUP BY s.student_id[, name] /** name required for SQL standard */
[ORDER BY name [ASC]] /** You'll probably want some kind of ordering */
[] indicate optional inclusions
I have a table of employees. Each employee has an employee id. Some employees have a supervisor field that links back to another employee's id. There are 10 employees, two of which are supervisors, each supervising 4 people. I am trying to get a list of all the employees and the number of other employees they supervise. So far I can only seem to get the supervisors and the number they supervise to show. This is my query:
SELECT s.employee_name, COUNT(*)
FROM employee e
join employee s on e.supervisor_id= s.employee_id
group by s.
order by s.employee_name;
I tried changing JOIN to RIGHT JOIN and it will now show me all 10 employees with the two supervisors shown as having 4 people they supervise but it shows all the others having no one to supervise as having 1 instead of 0. I'm sure it's something simple I am missing.
Sample Data:
employee_name, employee_name, supervisor_id,
'10111', 'Sydnee K. Stevens' NULL
'10870', 'Colton C. Rocha', '10111'
'11425', 'Astra V. Sharp','10111'
'12973', 'Melanie X. Rojas','10111'
'14451', 'Bethany Roman','10111'
'14597', 'Lydia Edwards', NULL
'16153', 'Selma Q. Conley', '14597'
'17730', 'Kristen B. Malone', '14597'
'17762', 'Barrett B. Bauer', '14597'
'18628', 'Shana Z. Flowers','14597'
We join your employee table with a select like we would join it with a real table. The select will consist of all supervisor_ids and the number of occurences in the supervisor_id field (records where supervisor_id is null will be ignored).
SELECT e.employee_id, e.employee_name, s.supervising
FROM employee e
LEFT JOIN (SELECT supervisor_id, count(*) as supervising
FROM employee
WHERE supervisor_id is NOT NULL
GROUP BY supervisor_id) AS s
ON(e.employee_id = s.supervisor_id)
Using LEFT or RIGHT JOIN COUNT(*) will always be at least 1. If using LEFT JOIN (RIGHT JOIN is confusing) you just need to count values from a right table column. COUNT(column) will ignore all rows with NULL in that column.
SELECT s.*, COUNT(e.supervisor_id) as num_supervised
FROM employee s
LEFT JOIN employee e on e.supervisor_id = s.employee_id
group by s.employee_id
order by s.employee_name;
I have 2 tables in a database person and order tables.
PERSON table:
PERSON_ID | NAME
ORDER table:
ORDER_ID | ORDER_NO | PERSON_ID
I need to display all the orders + a name of corresponding person if it exists, if not just order details.
So far I got up to query:
SELECT ORDER_ID, ORDER_NO, order.PERSON_ID, NAME
FROM person, order
WHERE person.PERSON_ID = order.PERSON_ID AND
person.FIRST_NAME IS NOT NULL;
Which gives me orders only if the name is available whereas I need to display all the orders despite the fact if name is available or not.
Any suggestions?
Yes, you can use LEFT JOIN for that:
SELECT o.order_id, o.order_no, o.person_id, p.name
FROM `order` o
LEFT JOIN person p
ON p.person_id = o.person_id AND p.FIRST_NAME IS NOT NULL
With LEFT JOIN if the name is null it will still give you the orders.
I have 3 tables I am wanting to join:
equipment
equipmentNo
assetNo
subCategoryNo
doNo
userNo
locationNo
isocNo
emergency
ariNo
makeNo
modelNo
serial
kitNo
purchaseDate
purchasePrice
comments
surplused
employees
userNo
fname
lname
doNo
emergency
subcategories
subCategoryNo
subCategory
categoryNo
replacementCycle
I need to get a set of data that includes all of the equipment information ordered by 'doNo' and 'lname' from employees which are linked by 'userNo' and the equipment 'subCategoryNo' needs to be only those from subcategories where 'categoryNo' equals 2.
I got this far before getting lost and confused.
SELECT equipment.*, employees.lname, employees.fname
FROM equipment
LEFT JOIN employees
USING (userNo)
ORDER BY equipment.doNo, employees.lname;
I could not figure out how to limit this to only 'categoryNo' 2;
SELECT equipment.*, employees.lname, employees.fname
FROM equipment
LEFT JOIN employees USING (userNo)
JOIN subcategories USING(subCategoryNo)
WHERE subcategories.categoryNo=2
ORDER BY equipment.doNo, employees.lname;