<div>
Please confirm that. 
 
Most recent MySQL I can find is version 8.0 and MariaDB is version 10.6.
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<div>
Sorry for the mistake <pre><code class="code-1 nolinks" id="code-20-43336248-1">Server version: 5.7.35-0ubuntu0.18.04.1 (Ubuntu)
</code></pre>
</div>

<div>
&gt;&gt;The result I would like to get is the following 
 
Are you sure that is what you are after? 
 
From what you described, I'm not seeing why you want: 

<pre><code class="code-1 nolinks" id="code-20-43336314-1">| 2 | Customer 2 | 3 | 2 | NULL | 2021-01-10 | 1200 |
</code></pre>
 
For two customers, I would expect only two offers: 

<pre><code class="code-1 nolinks" id="code-20-43336314-2">| 1 | Customer 1 | 2 | 1 | 1 | 2021-02-01 | 2000 |
+----+------------+----+------------+-------+-------------+--------+
| 2 | Customer 2 | 5 | 2 | 4 | 2021-03-10 | 2400 |
</code></pre>
 
If that is what you are after, here is what I came up with. 
 
Version 5.7 doesn't have cool things like Common Table Expressions or the ROW_NUMBER window function so you need to get hacky... 
 
I borrowed the ROW_NUMBER simulation code from: 
<a href="https://www.mysqltutorial.org/mysql-row_number/" rel="ugc">https://www.mysqltutorial.org/mysql-row_number/</a> 
 

<pre><code class="code-10 nolinks" id="code-20-43336314-3">select c.id, name, o.id, customerid, revid, mydate, amount
from customers as c
	join (
		select 
 		@row_number:=case
 		when @customer_id_v = customerid 
 		then 
 		@row_number + 1
 		else 
 		1
 		end as rn,
			id,
 		@customer_id_v:=customerid customerid,
			revid,
			mydate,
			amount
		from
 		offers,
 		(select @customer_id_v:=0,@row_number:=0) as t
		order by 
 		customerid, revid desc
	) as o on c.id=o.customerid and rn=1
;
</code></pre>
 
Complete working fiddle: 
<a href="https://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=4cfabde4c2d4b9968899bbd925662127" rel="ugc">https://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=4cfabde4c2d4b9968899bbd925662127</a>
</div>

Giorgio Todeschini

<div>
MySQL v8 and later supports the OVER clause and window functions such as ROW_NUMBER The "simulation" used by slightwv will work in practice (but strictly speaking this simulated approach isn't guaranteed to produce the expected results). Use of that simulation should be replaced once you do upgrade to v8 or later to use the supported functions e.g.: <pre><code class="code-10 nolinks" id="code-20-43336544-1">select c.id, name, o.id, customerid, revid, mydate, amount
from customers as c
inner join (
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; select 
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; row_number() over(partition by customerid, order by revid desc) as rn
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; , id
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; , revid
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; , mydate
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; , amount
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; from offers 
&nbsp; &nbsp; &nbsp; &nbsp; ) as o on c.id = o.customerid and o.rn = 1
</code></pre>As you can see the syntax here is a little easier on the eyes.
</div>

<div>
Dear slightwv, following your example, I found this way to solve the issue.<pre><code class="code-10 nolinks" id="code-20-43336591-1">SELECT * FROM customers
LEFT JOIN (
SELECT 
&nbsp; &nbsp;@row_number:=CASE
&nbsp; &nbsp;WHEN @off_id_v = offid 
&nbsp; &nbsp;THEN @row_number + 1
&nbsp; &nbsp;ELSE 1
&nbsp; &nbsp;END AS rn,
&nbsp; &nbsp;id,
&nbsp; &nbsp;@off_id_v:=offid offid,
&nbsp; &nbsp;customerid,
&nbsp; &nbsp;revid,
&nbsp; &nbsp;mydate,
&nbsp; &nbsp;amount
FROM (SELECT
&nbsp; &nbsp; &nbsp;&nbsp;@off_id:=CASE
&nbsp; &nbsp; &nbsp;&nbsp;WHEN revid IS NULL
&nbsp; &nbsp; &nbsp;&nbsp;THEN id
&nbsp; &nbsp; &nbsp;&nbsp;ELSE revid
&nbsp; &nbsp; &nbsp;&nbsp;END AS offid,
&nbsp; &nbsp; &nbsp;&nbsp;id,
&nbsp; &nbsp; &nbsp;&nbsp;customerid,
&nbsp; &nbsp; &nbsp;&nbsp;revid,
&nbsp; &nbsp; &nbsp;&nbsp;mydate,
&nbsp; &nbsp; &nbsp;&nbsp;amount
&nbsp; &nbsp;FROM
&nbsp; &nbsp; &nbsp;&nbsp;offers,
&nbsp; &nbsp; &nbsp;&nbsp;(SELECT @off_id:=0) AS t
&nbsp; &nbsp;ORDER BY offid, mydate DESC) AS tmp,
&nbsp; &nbsp;(SELECT @off_id_v:=0,@row_number:=0) AS t
) AS o ON o.customerid=customers.id AND rn=1
</code></pre> The result is the following that is exactly what I want. <pre><code class="code-1 nolinks" id="code-20-43336591-2">+----+------------+----+----+-------+------------+-------+------------+--------+
| id | name &nbsp; &nbsp; &nbsp; |&nbsp;rn | id | offid | customerid | revid | mydate &nbsp; &nbsp; | amount |
+----+------------+----+----+-------+------------+-------+------------+--------+
| &nbsp;1 | Customer 1 |&nbsp; 1 | &nbsp;2 | &nbsp; &nbsp; 1 | &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;1 | 1 | 2021-02-01 | &nbsp; 2000 |
+----+------------+----+----+-------+------------+-------+------------+--------+
| &nbsp;2 | Customer 2 | &nbsp;1 | &nbsp;3 | &nbsp; &nbsp; 3 | &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;2 | NULL &nbsp;| 2021-01-10 | &nbsp; 1200 |
+----+------------+----+----+-------+------------+-------+------------+--------+
| &nbsp;2 | Customer 2 | &nbsp;1 | &nbsp;5 | &nbsp; &nbsp; 4 | &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;2 | &nbsp; &nbsp; 4 | 2021-03-10 | &nbsp; 2400 |
+----+------------+----+----+-------+------------+-------+------------+--------+
</code></pre> What do you think? There is a better way to reach the same result? Thanks again.
</div>


I have 2 MySQL tables: 
Customers (master) 
+----+------------+
| ID | Name       |
+----+------------+
|  1 | Customer 1 |
+----+------------+
|  2 | Customer 2 |
+----+------------+
| .. | ...        |
+----+------------+
And Offers (details). 
+----+------------+-------+-------------+--------+
| ID | CustomerID | RevId | Date        | Amount |
+----+------------+-------+-------------+--------+
|  1 |          1 | NULL  | 2021-01-01  |   1000 |
+----+------------+-------+-------------+--------+
|  2 |          1 |     1 | 2021-02-01  |   2000 |
+----+------------+-------+-------------+--------+
|  3 |          2 | NULL  | 2021-01-10  |   1200 |
+----+------------+-------+-------------+--------+
|  4 |          2 | NULL  | 2021-02-10  |   2200 |
+----+------------+-------+-------------+--------+
|  5 |          2 |     4 | 2021-03-10  |   2400 |
+----+------------+-------+-------------+--------+
| .. |            |       |             |        |
+----+------------+-------+-------------+--------+
The RevId field, if not NULL,  indicate that the record of the Offers table is a revision of the offer with the ID=RevId in the same table.
For example, the second record of the table here above is a revision of the first one
Now, I would like to make a query like:
SELECT * FROM Customers LEFT JOIN Offers ON Customers.ID=Offers.CustomerID
but, when they are a revision, I want to select only the last record of the Offers table.
The result I would like to get is the following
+----+------------+----+------------+-------+-------------+--------+
| ID | Name       | ID | CustomerID | RevId | Date        | Amount |
+----+------------+----+------------+-------+-------------+--------+
|  1 | Customer 1 |  2 |          1 |     1 | 2021-02-01  |   2000 |
+----+------------+----+------------+-------+-------------+--------+
|  2 | Customer 2 |  3 |          2 | NULL  | 2021-01-10  |   1200 |
+----+------------+----+------------+-------+-------------+--------+
|  2 | Customer 2 |  5 |          2 |     4 | 2021-03-10  |   2400 |
+----+------------+----+------------+-------+-------------+--------+
I'm not new in SQL language, I tried a very lot of ways but I can't able to solve this issue
Could anyone please help me?
Thanks in advance

Join selecting only the the last row on the details table

SQL (Structured Query Language) is designed to be used in conjunction with relational database products as of a means of working with sets of data. SQL consists of data definition, data manipulation, and procedural elements. Its scope includes data insert, query, update and delete, schema creation and modification, and data access control.

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