How to use subquery with group by in mysql? - php

Question

I have one table with multiple rows for particular users.i am having
data with many years like 2019,2018,2016 and more. i have two
scenarios:
1.i want data of particular INVOICE_YEAR.
2,but same time i want FIRST DATE OF INVOICE GENERATED FOR PARTICULAR
USER.
my sql query :
$yearOf this is dynamic year input variable.
$yearOf = 2019;
$Records = "SELECT MIN(inv.INVOICE_DATE) AS MIN_INVOICE_DATE
FROM invoices as inv
LEFT JOIN customers as cm ON cm.CUSTOMER_ID = inv.CUSTOMER_ID
where inv.INVOICE_YEAR IN (".$yearOf.")
group by inv.CUSTOMER_ID ORDER BY cm.CUSTOMER_NAME ASC";
As you can see my query if i want data of all users where INVOICE_YEAR IN ('2019').My first condition will satisfied i will get all data of users where INVOICE_YEAR = 2019.
But same time i want first invoice date so for this i used MIN(inv.INVOICE_DATE) but as i used
where inv.INVOICE_YEAR IN (".$yearOf.") this in where condition so it gives me first invoice date of particular year.
But i want first invoice date from whole table for all users.
I tried with subquery but it shows me error of Subquery returns more than 1 row
My query with subquery:
$Records = "SELECT
(
SELECT MIN(inv.INVOICE_DATE) AS MIN_INVOICE_DATE FROM invoices AS inv GROUP BY inv.CUSTOMER_ID) AS MIN_INVOICE_DATE
FROM invoices as inv
LEFT JOIN customers as cm ON cm.CUSTOMER_ID = inv.CUSTOMER_ID
where inv.INVOICE_YEAR IN (".$yearOf.")
group by inv.CUSTOMER_ID ORDER BY cm.CUSTOMER_NAME ASC";
For eg:
3 customers are there 101,102,103
data in table is like:
id | customer_id | invoice_date | invoice_year
1 | 101 | 2019-01-01 | 2019
2 | 101 | 2016-01-01 | 2016
3 | 101 | 2017-01-01 | 2017
4 | 101 | 2016-01-02 | 2016
5 | 102 | 2019-01-02 | 2019
6 | 103 | 2018-01-02 | 2018
7 | 103 | 2019-01-07 | 2019
8 | 102 | 2015-01-02 | 2015
As i Request query to get data of INVOICE_YEAR 2019 with first invoice date of particular user so it should give output like :
id | customer_id | invoice_date | invoice_year | min_invoice_date
1 | 101 | 2019-01-01 | 2019 | 2016-01-01
5 | 102 | 2019-01-02 | 2019 | 2015-01-02
7 | 103 | 2019-01-07 | 2019 | 2019-01-07
With as i want first invoice date IN COLUMN MIN_INVOICE_DATE of all users.
But it shows me data like :
id | customer_id | invoice_date | invoice_year | min_invoice_date
1 | 101 | 2019-01-01 | 2019 | 2019-01-01
5 | 102 | 2019-01-02 | 2019 | 2019-01-02
7 | 103 | 2019-01-07 | 2019 | 2019-01-07

Answer 1

You want to pull out the earliest invoice for each customer in 2019, along with the date of their earliest invoice within the whole table.
In MySQL 8.0, you can solve this using window functions:
SELECT id, customer_id, invoice_date, invoice_year, min_invoice_date
FROM (
SELECT
t.*,
ROW_NUMBER() OVER(PARTITION BY customer_id, invoice_year ORDER BY invoice_date) rn,
MIN(invoice_date) OVER(PARTITION BY customer_id) min_invoice_date
FROM mytable t
) x
WHERE invoice_year = 2019 AND rn = 1
In earlier versions, you can:
JOIN the table with a query that computes the overall minimum invoice_date per customer
use a correlated subquery with a NOT EXISTS condition to filter on the earliest invoice par customer in 2019
Query:
SELECT t.id, t.customer_id, t.invoice_date, t.invoice_year, m.min_invoice_date
FROM mytable t
INNER JOIN (
SELECT customer_id, MIN(invoice_date) min_invoice_date
FROM mytable
GROUP BY customer_id
) m ON m.customer_id = t.customer_id
WHERE
t.invoice_year = 2019
AND NOT EXISTS (
SELECT 1
FROM mytable t1
WHERE t1.invoice_year = 2019
AND t1.customer_id = t.customer_id
AND t1.invoice_date < t.invoice_date
)
In this demo on DB Fiddle, both queries return:
| id | customer_id | invoice_date | invoice_year | min_invoice_date |
| --- | ----------- | ------------ | ------------ | ---------------- |
| 1 | 101 | 2019-01-01 | 2019 | 2016-01-01 |
| 5 | 102 | 2019-01-02 | 2019 | 2015-01-02 |
| 7 | 103 | 2019-01-07 | 2019 | 2018-01-02 |

Answer 2

I guess you want the to see the details of the first (earliest-dated) invoice for each customer in each calendar year. You want that result filtered to cover only one year.
So, start with a subquery to find the date of the first invoice for every customer in each calendar year. (https://www.db-fiddle.com/f/bmBZ14Vr9Re6ahpfs2FF2X/0)
SELECT MIN(invoice_date) first_invoice_date,
YEAR(invoice_date) calendar_year,
customer_id
FROM invoices
GROUP BY YEAR(invoice_date), customer_id
Then retrieve the detail for those invoices by JOINing that subquery to your original invoices table. (https://www.db-fiddle.com/f/bmBZ14Vr9Re6ahpfs2FF2X/1)
SELECT invoices.*
FROM invoices
JOIN (
SELECT MIN(invoice_date) first_invoice_date,
YEAR(invoice_date) calendar_year,
customer_id
FROM invoices
GROUP BY YEAR(invoice_date), customer_id
) firsts
ON invoices.customer_id = firsts.customer_id
AND invoices.invoice_date = firsts.first_invoice_date
ORDER BY invoices.customer_id,
invoices.invoice_year,
invoices.invoice_date
Then, throw in WHERE invoices.invoice_year = 2019 to get just the year you want.
Notice that your invoice_year column is unnecessary, since it can always be computed from YEAR(invoice_date). You should consider getting rid of it.