I have two tables (orders, order_lines) with a one to many relationship. The orders table will have one row and the order_lines table will have one row for each product attached to the order.
ORDERS
+----------+-------------+---------------+
|order_num | total_order | shipping_total|
+----------+-------------+---------------+
| 12345 | 75.00 | 15.00 |
+----------+-------------+---------------+
ORDER_LINES
+----------+-------------+-------+
| order_num| volume | price |
+----------+-------------+-------+
| 12345 |10 |25.00 |
+----------+-------------+-------+
| 12345 |20 |25.00 |
+----------+-------------+-------+
| 12345 |20 |25.00 |
+----------+-------------+-------+
I would like to sum the total_order & shipping_total columns in the ORDERS table and the volume & price columns in the ORDER_LINES table.
Query currently not working:
SELECT b_orders.bill_country, b_orders.ship_country
, SUM(volume) AS v, SUM(price) AS pt
, SUM(shipping_total) AS st, SUM(total_order) AS tot_o
FROM b_orders
JOIN b_order_lines
ON b_orders.order_num = b_order_lines.order_num
WHERE DATE(b_orders.order_date) BETWEEN '2012-04-02' AND '2012-04-06'
AND ship_country = 'USA'
AND b_order_lines.price > 0;
This query is returning the wrong values. I kind of know why but have no idea how to write the proper query...please help.
When you use an aggregate function line SUM, you need to tell the SQL server what columns are going to be used to GROUP BY for doing the sums.
So if you add the following to the bottom of your query:
GROUP BY
b_orders.bill_country, b_orders.ship_country
This will tell it to give you the sum for each unique combination of bill_country, ship_country in the data. Is this what you are looking for?
If bill_country and ship_country are going to be included in a query that also has aggregate functions like SUM, you need to group on them. I suspect all you need is to add this line at the end:
GROUP BY b_orders.bill_country, b_orders.ship_country;
If you don't need to see them, you can remove them from the SELECT statement and omit the grouping; you don't have to select columns to include them in the WHERE clause.
Related
I have a table that contains information about an item, and another table that references the owner of that item, like so:
baseItem
--------
itemID | 1 | 2 | 3 | 4 |
itemSize | 5 | 1 | 5 | 3 |
itemCost | 100 | 50 | 1 | 99 |
itemOwner
--------
ownerID | 1 | 1 | 3
itemID | 1 | 4 | 2
What I'm after are the SUMS of itemSize and itemCOST based on the owner. I've looked around but none of the answers I've seen make sense? Here's the best I could come up with, which clearly isn't working:
SUM itemCost FROM baseItem.itemCost LEFT JOIN itemID ON itemOwner.itemid = baseItem.itemid
SELECT ownerId, sum(itemCost) as OwnerCost, sum(itemSize) as OwnerSize
FROM itemOwner
LEFT JOIN baseItem
ON itemOwner.itemid = baseItem.itemid
GROUP BY ownerId
A SELECT statement lists which fields you want to read from the table; in this case you want two values: the id of the owner, and the sum of the values of the items they own. However, since you're using sum (an aggregate function), you must GROUP your elements over some parameter. In this case, you want to group them by ownerId.
A FROM clause references a table; you can start with either baseItem or itemOwner, it makes no difference in this case. You can think of LEFT JOIN as a cartesian product that creates a new table, which contains every element from the cartesian product of both, filtered by the ON clause. However will always have all the items in the left table which is itemOwner and when there are no matching rows in the baseItem all the fields will be NULL. The SUM function will act as if those are 0s and should return you a 0 sum for non matching rows in the baseItem table.
Maybe it is not working because it is invalid SQL statement. Try following code
SELECT SUM(baseItem.itemCost) FROM baseItem
LEFT JOIN itemOwner ON itemOwner.itemId = baseItem.itemId
take the case you have 2 table, for example tbCostumers and tbOrders.
I would like to display a summary list with all costumers, related orders and display them with a paginator.
Doing a join I can extract costumers list and all orders for each costumer, the result is something like:
idCostumer | name | ... | idProduct | productName | price | ...
Where the first n columns are all equal if the costumer has more than 1 order. So I can have:
1 | will | ... | 12 | product1 | 123 | ...
2 | bill | ... | 23 | product2 | 321 | ...
2 | bill | ... | 24 | product3 | 231 | ...
And so on
I'm trying to use LIMIT to extract only n records and using them with a paginator.
First question: if a costumer has more than 1 order, with this query I'll see n records, equal in the first column (id, name, ... and other costumer info) but different at the end, where there are products info. Is this 'correct'? Is there another way to extract this informations?
Second question: if I do that and I use a LIMIT, I could "cut" the result table between 2 (or more) records that represent the same customer; so, for example in the small table above, if I limit with 2 the third row will be lost, even if it's part of the row above, because is just another order of the same costumer.
I would like to limit the number of different idCostumer, in order to take exactly n costumers, even if they appear more than 1 times in the result table. Something like n different idCostumer, no matter if they are repeated.
Is this possible?
I hope it's clear, it was not easy to explain what I would like to achieve :)
Thank you!
You might want to have something like this:
SELECT * FROM (
(SELECT * FROM tbCustomers LIMIT 3) AS c
INNER JOIN tbOrders AS o ON o.customer = c.idcustomer
);
You can substitute the first asterisk with named columns and only receive your desired columns in the order you prefer (ie: SELECT c.name, o.price FROM...) .
Hope this works for you!
EDIT: changing the value of the LIMIT clause changes the number of the picked customers, of course.
EDIT 2: As Alvaro Pointed out, you'll probably need an order clause in the tbCustomers query.
I've searched for a few hours now, but couldn't find relative solution to a specific algorithm I am working on. To simplify the obstacle, I would like to present the information in just one table.
_____________________________________
| User | Item | price | qty |
-------------------------------------
| Annie | Dress | 80 | 1 |
| Bob | Jeans | 65 | 3 |
| Cathy | Shoes | 60 | 4 |
| David | Shirts | 40 | 6 |
| Annie | Shoes | 60 | 2 |
| Bob | Shirts | 55 | 2 |
| Cathy | Jeans | 65 | 1 |
| David | Ties | 20 | 5 |
-------------------------------------
Problem # 1: Show users whose total price for shopping at the store is 300 or more and quantity of their purchase is less than or equal to 3. These shoppers will be mailed a coupon for $40.
Problem # 2: Show users whose total qty is greater than or equal to 7 and the total for price is 275 or more. These shoppers will be mailed a coupon for $20.
The rows within the table are not transaction specific. The table can represent separate transactions within a month. We're just trying to find certain returning customers who we would like to reward for shopping with us.
I'm not sure if this can be done only via MySQL, or if I need to have separate queries and store rows into arrays and compare them one by one.
What I have tried so far are the followings:
SELECT * FROM table where SUM(price) as Total >= 300 AND SUM(qty) <=3;
I've also tried the following after the research:
SELECT SUM(price) as Total FROM table WHERE SUM(qty) <=3;
I keep getting syntax errors in MySQL shell. You don't have to solve the problems for me, but if you can guide me through the logic on how to solve the problems, I'd appreciate it very much.
Lastly I'd like to ask once, can I solve this with only MySQL or do I need to store the rows into PHP arrays and compare each indexes?
You can't use an aggregate function in the WHERE clause, you have to use HAVING. WHERE operates on individual rows during the selection, HAVING operates on the final results after aggregating.
SELECT *, SUM(price*qty) as Total
FROM table
GROUP BY user
HAVING Total >= 300 AND SUM(qty) <= 3
SUM is an aggregate function, meaning it applies to a group of clubbed rows. S say i am grouping the table data based on NAME then sum function would sum all the price of one NAME.
Having said this, if you think logically it would not make any sense to put the sum(price) in a WHERE clause because where clause would not know which SUM(PRICE) for which NAME to operate on(where clause operates only after a temporary view has been generated).
So we have the HAVING clause in SQL. This is used to compare the results of aggregrate function at each step of aggregation.
Consider it like this:
In where clause, when the ANNIE row from your DB is returned, it does not know what SUM(PRICE) means.
While in HAVING clause the SUM(PRICE)>300 condition is executed only when SQL has finished grouping all the ANNIE data into one group and calculated the SUM(PRICE) for her.
For question 1:
SELECT USER, SUM(PRICE)
FROM table
GROUP BY user
HAVING SUM(PRICE) >= 300 AND SUM(QTY) <= 3
For Question 2:
SELECT USER, SUM(PRICE)
FROM table
GROUP BY user
HAVING SUM(PRICE) >= 275AND SUM(QTY) >=7
we are working with a table that has a column for every SKU ordered by a customer. So if a particular SKU has been ordered by five different customers, it will appear 5 times. Also, some customers may order 2 of that SKU so in the 'number_ordered' column next to it, there will be a 2. I'm not so good at drawing this out in words so I'll give an example of what the database looks like.
+------------+-----------------+
| item_SKU | Number Ordered |
+------------+-----------------+
| SKU001 | 3 |
| SKU001 | 2 |
| SKU002 | 15 |
| SKU003 | 1 |
+------------+-----------------+
How can I times the sku by the number ordered and then add them all together in MySQL. I need to put it into PHP but I can do that if i get some hints on how to do this.
Cheers
Use this to get the total ordered item of same item_SKU and use it in PHP. If you need the order count also, use COUNT() function
SELECT item_SKU, SUM(Number_Ordered) as Total_Ordered_Item, COUNT(Number_Ordered) as Total_Order_Count
FROM table
GROUP BY item_SKU
Try using count' with group by sku name (SKU001). it returns an output with the count of different sku's.
try below SQL query
1.Sum of Orders
select item_SKU,sum(number_ordered) as orders from `table Name` group by item_SKU order by item_SKU
output
item_SKU | Number Ordered
SKU001 | 5
SKU002 | 15
SKU003 | 1
2. Query to count no of orders
select item_SKU,count(number_ordered) as orders from `table Name` group by item_SKU order by item_SKU
output
item_SKU | Number Ordered
SKU001 | 2
SKU002 | 1
SKU003 | 1
Try this query:
SELECT item_SKU,
COUNT(item_SKU) CNT,SUM(Number_Ordered) TOTAL
FROM TABLE1 GROUP BY item_SKU;
SQL Fiddles
Select item_SKU, SUM(NumberOrdered) As total from [table name] group by item_SKU
Try this one. It gives you total no of occurrence as 'COUNT' and sum of number_ordered as 'SUM_NUM_ORDER'
SELECT item_SKU,COUNT(item_SKU) as COUNT,SUM(Number_Ordered) as SUM_NUM_ORDER FROM TABLE1 GROUP BY item_SKU;
DEMO: http://sqlfiddle.com/#!2/9ec2b9/10
i have a form that has a multiple select drop down. a user can select more than one options in the select. the name of the select is array[]; using php I call implode(",",$array)
in mysql db, it stores the field as a text in this format "places"= "new york, toronto, london" when i want to display these fields i explode the commas.
I am trying to run a report to display the places. here is my select:
"select * from mytable where db.places .. userSelectedPlaces"
how can i check toronto in lists of "places" that user selected? note "places" in the db might be either just "toronto" or it might be comma separated lists of places like "ny, toronto, london, paris, etc".
If it is possible, you would be much better off using another table to hold the places that the user has selected. Call it SelectedPlaces with columns:
mytable_id - To join back to the table in your query
place - EG: "Toronto"
Then you can run a simple query to figure out if Toronto has been selected:
SELECT *
FROM mytable m
INNER JOIN SelectedPlaces sp ON sp.mytable_id = m.id
WHERE sp.place = 'Toronto'
If I understand you correctly, your database design is just wrong. Try reading about it more. Generally, in good design you should not have lists of values as one field in database and you should introduce new table for it.
But if you want to do it this way, you can use strcmp function.
If i understood correctly, this should work:
WHERE DB.PLACES LIKE '%TORONTO%'
but as other users said, its not a nice thing to have denormalized tables.
To directly answer your question, your query needs to look something like this
SELECT *
FROM mytable
WHERE places LIKE( '%toronto%' )
But, be aware, that LIKE() is slow.
To indirectly answer your question, your database schema is all wrong. That is not the right way to do a M:N (many-to-many) relationship.
Imagine instead you had this
mytable place mytable_place
+------------+ +----------+----------+ +------------+----------+
| mytable_id | | place_id | name | | mytable_id | place_id |
+------------+ +----------+----------+ +------------+----------+
| 1 | | 1 | new york | | 1 | 1 |
| 2 | | 2 | toronto | | 1 | 2 |
| 3 | | 3 | london | | 1 | 3 |
+------------+ +----------+----------+ | 2 | 2 |
| 3 | 1 |
| 3 | 3 |
+------------+----------+
The table mytable_places is what's called a lookup table (or, xref/cross-reference table, or correlation table). Its only job is to keep track of which mytable records have which place records, and vice versa.
From this example we can see that The 1st mytable record has all 3 places, the 2nd has only toronto, and the 3rd has new york and london.
This opens you up too all sorts of queries that would be difficult, expensive, or impossible with your current design.
Want to know how many mytable records have toronto? No problem
SELECT COUNT(*)
FROM mytable_place x
LEFT JOIN place p
ON p.place_id = x.place_id
WHERE p.name = 'toronto';
How about the number of mytable records per place, sorted?
SELECT p.name
, COUNT(*) as `count`
FROM mytable_place x
LEFT JOIN place p
ON p.place_id = x.place_id
GROUP BY p.place_id
ORDER BY `count` DESC, p.name ASC
And these are going to be much faster than any query using LIKE since they can use indexes on columns such as place.name.