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What are the benefits of only_full_group_by mode?

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sql

mysql

I updated mysql and I went from MySQL Version 5.6.17 to version 5.7.14

Since I have errors on my sql queries

Indeed, many of my queries look like this:

SELECT count (id) as nbr, lic from prep WHERE key = '18' 

And I have this error:

1140 - In aggregated query without GROUP BY, expression #2 of SELECT list contains nonaggregated column 'operator.preparation.orig_lic'; this is incompatible with sql_mode=only_full_group_by

After some research, I learn that Mysql 5.7.14 activates ONLY_FULL_GROUP_BY by default

Why is it enabled by default?

What is the best solution (for performance)? Disable ONLY_FULL_GROUP_BY or add a 'group by' on my query?

Thank you

like image 548
Rocstar Avatar asked Aug 03 '17 12:08

Rocstar


People also ask

What is the use of Only_full_group_by?

If ONLY_FULL_GROUP_BY is disabled, a MySQL extension to the standard SQL use of GROUP BY permits the select list, HAVING condition, or ORDER BY list to refer to nonaggregated columns even if the columns are not functionally dependent on GROUP BY columns. This causes MySQL to accept the preceding query.

What is Nonaggregated column MySQL?

If the ONLY_FULL_GROUP_BY SQL mode is enabled (which it is by default), MySQL rejects queries for which the select list, HAVING condition, or ORDER BY list refer to nonaggregated columns that are neither named in the GROUP BY clause nor are functionally dependent on them.


2 Answers

only_full_group_by = on tells MySQL engine: Do not apply GROUP BY when you have doubt about what results to show and throw an error. Only apply it if Group By specifically tells you what to do. i.e. when the Group By is full and perfect!

only_full_group_by = off tells MySQL engine: always apply GROUP BY and if you have doubt about what results to choose, just pick one randomly!

You don't have to turn it off if you use GROUP BY properly!

Example:

Table: users

 id   |  name ----------------   1      ali   2      john   3      ali 

When you use GROUP BY on the name column:

SELECT * FROM users GROUP BY name; 

There are two possible results:

  1      ali   2      john      

OR

  2      john   3      ali 

MYSQL does not know what result to choose! Because there are different ids but both have name=ali.

Solution1:

only selecting the name field:

SELECT name FROM users GROUP BY name; 

result:

  ali   john      

This is a perfect solution. removing columns that makes GROUP BY confused. This means you know what you're doing. Usually, you do not need
those columns, but if you need them, go to Solution3!

Solution2:

Turning off only_full_group_by. MYSQL will pick one of the two possible results RANDOMLY!! (It's ok if you do not really care what id it will choose, but remember to turn it on immediately after your query to prevent unexpected behaviors in future groupBys)

Solution3

Use an Aggregate function like MIN(), MAX() to help MYSQL to decide what it must choose.

For example:

SELECT MAX(id), name FROM users GROUP BY name; 

result:

  2      john        3      ali 

It will choose the ali row which has the maximum id.

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Ahmad Mobaraki Avatar answered Oct 01 '22 21:10

Ahmad Mobaraki


The "best" solution is to do the correct thing and fix your query by adding a group by, rather than override the error being thrown. If you override the error with ONLY_FULL_GROUP_BY the error your experiencing will go away but you'll likely experience two new errors as a result of doing so:

  1. Unexpected results of including an aggregated value with non-aggregated values, the problem your error is trying to prevent.

  2. Inability to execute your query on other environments. If you ever need to switch settings or give your code to someone else not using this database, the query will throw the error again. If you get into a habit of overriding the error or other errors, your code could become unusable to others and severely cripple the usefulness of it.

In general, if you are receiving an error, fix it rather than just telling the compiler/optimizer to ignore it.

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yanman1234 Avatar answered Oct 01 '22 22:10

yanman1234