MariaDB Select Min-Date (earliest) from multiple columns

Question

I need to do a select within a MariaDB, where I have one row per customer and the earliest date out of three different tables with action datetime values.

Example Tables:

Main Table

customer-id	Column_X
First	Things
Second	Things

Table One

customer-id	Date
First	Date
Second	Date

Table Two

customer-id	Date
First	Date
Second	Earliest Date (Table2)

Table Three

customer-id	Date
First	Earliest Date (Table3)
Second	Date

My aim is to have the earliest date out of the three columns in the other tables in one column within the select.

What I tried to do is this:

SELECT main.customer-id , main.Column_X 
    (SELECT LEAST(C) FROM (VALUES ((table1.date) , (table2.date), (table3.date)) AS C) AS First_Action
FROM main_table main
LEFT JOIN table_one table1 ON table1.cutomer-id = main.customer-id
LEFT JOIN table_two table2 ON table2.cutomer-id = main.customer-id
LEFT JOIN table_three table3 ON table3.cutomer-id = main.customer-id
GROUP BY main.customer-id;

Unfortunatly, I don't get any results just an error message.

So the resulting table should look something like this:

Result

customer-id	Column_X	First_Action
First	Things	Earliest Date (Table 3)
Second	Things	Earliest Date (Table 2)

I just started working with SQL statments and therefore have basically no experience. Help would be much appreciated!

Many Greetings Chris

Answer 1

It seems you simply need a LEAST function -

SELECT main.customer-id , main.Column_X,
       LEAST(table1.date, table2.date, table3.date) AS First_Action
  FROM main_table main
  LEFT JOIN table_one table1 ON table1.cutomer-id = main.customer-id
  LEFT JOIN table_two table2 ON table2.cutomer-id = main.customer-id
  LEFT JOIN table_three table3 ON table3.cutomer-id = main.customer-id;