I cant select data from database .
My table structrure is given below
customer table
id name
10 geetha
customer country table
id cust_id country
1 10 6
2 10 16
I got the result like these way
customer name country
geetha 6
geetha 16
But i want to get the one customer data only one time ie with out repeating.
customer name country
geetha 6
my query is
SELECT customer.name,customer.id,customer_country.country_id, customer_country.cust_id
FROM customer
INNER JOIN customer_country on customer.id = customer_country.cust_id
If you have duplicates in the customer_country table, then you need to choose one of them. Here is one method using max():
select c.name, max(cc.country_id)
from customer c inner join
customer_country cc
on c.id = cc.cust_id
group by c.name;
If you want all of them in a list, use group_concat():
select c.name, group_concat(cc.country_id) as countries
from customer c inner join
customer_country cc
on c.id = cc.cust_id
group by c.name;
For first record only, apply to the end: limit 1
SELECT customer.name,customer.id,customer_country.country_id,
customer_country.cust_id from customer
inner join customer_country on customer.id= customer_country.cust_id limit 1
try this i add distinct before customer_country.cust_id
SELECT customer.name,customer.id,customer_country.country_id, distinct customer_country.cust_id
FROM customer
INNER JOIN customer_country on customer.id = customer_country.cust_id
Related
Here i have 3 tables name A,B,C respectively and i want to join all the tables and fetch out the results
In order to get the desired output i wrote my code like this
SELECT A.date as d_date,B.agent_name,
(SELECT SUM(B.profit) FROM B WHERE A.id = B.bill_id) AS total_profit,
SUM(C.total_price) AS t_price,SUM(C.total_dc) AS t_dc
FROM A LEFT JOIN B ON A.id=B.bill_id
LEFT JOIN C ON C.data_id=B.id
WHERE DATE(A.date) BETWEEN '{$start_date}' AND '{$end_date}'
AND A.customerid=406
GROUP BY Date(A.date),A.customerid
ORDER BY A.id;
The problem is am getting Purchase value as the first value of the profit column from the table B.
i want my desired output to be like this
Name Date Purchase t_price t_dc
Ned 2019-07-26 210.60 80 40
but am getting like this
Name Date Purchase t_price t_dc
Ned 2019-07-26 15.60 80 40
here is the demo http://sqlfiddle.com/#!9/c85a910/3
The problem here is, Table C have 2 rows and both having data_id as 67159. So when you will join it with table B it will count the profit for bill_id 67159 twice. You have to use one condition to pick only 1 row. I have updated your query to -
SELECT A.date as d_date,B.agent_name, SUM(B.profit) AS total_profit,
SUM(C.total_price) AS t_price,SUM(C.total_dc) AS t_dc
FROM A LEFT JOIN B ON A.id=B.bill_id
AND A.customerid = B.user_id
LEFT JOIN C ON C.data_id=B.id
WHERE DATE(A.date) BETWEEN '2019-07-26' AND '2019-07-26'
AND A.customerid=406
GROUP BY Date(A.date),B.agent_name
ORDER BY A.id;
This query is giving total_profit as 226.2.
I'm joining two tables to display car brands. Here is the structure:
SELECT DISTINCT
b.title
FROM
brands as b
INNER JOIN items as i
ON i.brand_id = b.id
WHERE i.status = 1
ORDER BY COUNT(i.brand_id) DESC;
The above only produces one record. If I remove "ORDER BY COUNT(i.brand_id) DESC;" it displays all the records correctly.
I would like to sort result based on number of vehicles under each brand category. So for example if bmw category has the most car listed under, it should be the first one.
SELECT b.title
FROM brands as b
INNER JOIN items as i
ON i.brand_id = b.id
WHERE i.status = 1
GROUP BY b.title
ORDER BY COUNT(i.brand_id) DESC;
This should work for you.
I would use
SELECT b.title, count(i.brand_id)
FROM items i
LEFT JOIN brands as b
ON b.id = i.brand_id
WHERE i.status = 1
GROUP BY i.brand_id
ORDER BY COUNT(i.brand_id) DESC;
Your concern is the amount of cars in inventory. You want to break them down by how many of each brand you have. So, you're mainly concerned with the items tables and only need the brands table to get the information stored in the items table (brand name). Lastly, in order to get aggregate the number of brands of each, you must let MySQL know what you want in the GROUP BY.
I haven't used an EXPLAIN, but I would think the bottom query is more efficient.
You could join brands on an aggregate query:
SELECT b.title
FROM brands b
INNER JOIN (SELECT brand_id, COUNT(*) AS cnt
FROM items
WHERE status = 1
GROUP BY brand_id) i ON i.brand_id = b.id
ORDER BY cnt DESC;
I have this tables
ORDERS
orders_id
order_article_id
order_invoice_id
order_customer_id
order_qty
ARTICLES
articles_id
article_name
article_qty
article_price
article_amount
CUSTOMERS
customers_id
customer_name
customer_position
customer_office
And SQL Join Table
$sql="SELECT customer_name, article_name, orders_id FROM orders
LEFT JOIN articles ON order_article_id = articles_id
LEFT JOIN customers on order_customer_id = customers_id";
From this query I need to get AMOUNT.
Amount is example
USER: MICHAEL
AMOUNT: ORDER ID = order_article_id + order_qty;
Is it possible to do that in MySQL or i need some addition PHP code?
You can do it in MySQL, using a GROUP BY query and a SUM() aggregate function:
SELECT
c.customer_name,
SUM(a.article_price*a.article_quantity) AS amount
FROM
orders o LEFT JOIN articles a
ON o.order_article_id = a.articles_id
LEFT JOIN customers c
ON o.order_customer_id = c.customers_id
GROUP BY
c.customers_id,
c.customer_name
you might want to obtain the AMOUNT per order, then you need to group by also for the order_id:
GROUP BY
o.orders_id,
c.customers_id,
c.customer_name
Basically I have a load of product information on Table A. This include the product_id which is the common id over both tables. On Table B I have a list of votes which include product_id, username, thevote(could be +1 or -1).
So basically I want to have a table of 'Table A' with a additional column containing the SUM of all the votes for that product_ID. I am sure there is an easy way to do this. Which i think is using 'right join'.
I still want it to list all the products in Table A regardless if they have a single +1 or -1 vote.
Many thanks in advance peoples!
You could use group by statement and left join like:
select productName, sum(vote) as productVoteSum
from `products` p
left join `products_votes` pv on p.id = pv.productId
where productName like '%chocolat%'
group by p.id
order by productName;
You can do a LEFT JOIN (that goes through all the information on tableA and puts the SUM in the ones that have the matching product_id on tableB), then you add the SUM(b.thevote) and group by the remaining columns
SELECT a.product_id,a.productName,SUM(b.thevote)
FROM tableA a
LEFT JOIN tableB b ON a.product_id = b.product_id
GROUP BY a.product_id, a.productName
SELECT a.*, SUM(b.votes) FROM TableA a LEFT JOIN TableB b ON a.prouct_id=b.product_id
GROUP BY a.product_id
I have 3 tables:
shops, PRIMARY KEY cid,zbid
shop_items, PRIMARY KEY id
shop_inventory, PRIMARY KEY id
shops a is related to shop_items b by the following: a.cid=b.cid AND a.zbid=b.szbid
shops is not directly related to shop_inventory
shop_items b is related to shop_inventory c by the following: b.cid=c.cid AND b.id=c.iid
Now, I would like to run a query which returns a.* (all columns from shops). That would be:
SELECT a.* FROM shops a WHERE a.cid=1 AND a.zbid!=0
Note that the WHERE clause is necessary.
Next, I want to return the number of items in each shop:
SELECT
a.*,
COUNT(b.id) items
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid
As you can see, I have added a GROUP BY clause for this to work.
Next, I want to return the average price of each item in the shop. This isn't too hard:
SELECT
a.*,
COUNT(b.id) items,
AVG(COALESCE(b.price,0)) average_price
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid
My next criteria is where it gets complicated. I also want to return the unique buyers for each shop. This can be done by querying shop_inventory c, getting the COUNT(DISTINCT c.zbid). Now remember how these tables are related; this should only be done for the rows in c which relate to an item in b which is owned by the respective shop, a.
I tried doing the following:
SELECT
a.*,
COUNT(b.id) items,
AVG(COALESCE(b.price,0)) average_price,
COUNT(DISTINCT c.zbid)
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
LEFT JOIN shop_inventory c ON c.cid=b.cid AND c.iid=b.id
WHERE a.cid=1
GROUP BY b.szbid,b.cid
However, this did not work as it messed up the items value. What is the proper way to achieve this result?
I also want to be able to return the total number of purchases made in each shop. This would be done by looking at shop_inventory c and adding up the c.quantity value for each shop. How would I add that in as well?
Try this solution:
SELECT a.*,
COALESCE(b.item_cnt, 0) AS item_cnt,
COALESCE(b.avg_price, 0) AS avg_price,
COALESCE(b.buyer_cnt, 0) AS buyer_cnt
FROM shops a
LEFT JOIN (
SELECT a.cid,
a.szbid,
COUNT(*) AS item_cnt,
AVG(a.price) AS avg_price,
b.buyer_cnt
FROM shop_items a
LEFT JOIN (
SELECT cid,
iid,
COUNT(DISTINCT zbid) AS buyer_cnt
FROM shop_inventory
WHERE cid = 1
GROUP BY cid,
iid
) b ON a.cid = b.cid AND a.id = b.iid
WHERE a.cid = 1 AND
a.szbid <> 0
GROUP BY a.cid,
a.szbid
) b ON a.cid = b.cid AND a.zbid = b.szbid
WHERE a.cid = 1 AND
a.zbid <> 0
Instead of COUNT(DISTINCT c.zbid) + LEFT JOIN shop_inventory you could write a subselect:
SELECT
a.*,
COUNT(b.id) items,
AVG(COALESCE(b.price,0)) average_price,
( SELECT COUNT(DISTINCT c.zbid)
FROM shop_inventory c
WHERE c.cid=b.cid AND c.iid=b.id
)
FROM shops a
LEFT JOIN shop_items b ON b.cid=a.cid AND b.szbid=a.zbid
WHERE a.cid=1
GROUP BY b.szbid,b.cid