Combine Multiple child rows into one row MYSQL

MysqlSelectJoin

Mysql Problem Overview


Thanks in advance, I just can't seem to get it!

I have two tables

Ordered_Item

ID | Item_Name
1  | Pizza
2  | Stromboli

Ordered_Options

Ordered_Item_ID | Option_Number | Value
1               43         Pepperoni
1               44         Extra Cheese
2               44         Extra Cheese

What I am looking to output is a mysql query is something to this effect

Output

ID | Item_Name | Option_1 | Option_2
1    Pizza       Pepperoni  Extra Cheese
2    Stromboli     NULL     Extra Cheese

I have tried numerous options most ending in syntax error, I have tried group_concat but thats not really what I am looking for. I have a crude example below of what I think might be a start. I need the options to be in the same order every time. And in the program where the info is collected there is no way to reliable ensure that will happen. Is it possible to have them concatenate according to option number. Also I know that I will never have over 5 options so a static solution would work

Select Ordered_Items.ID,
    Ordered_Items.Item_Name,
FROM Ordered_Items
    JOIN (SELECT Ordered_Options.Value FROM Ordered_Options Where Option_Number = 43) as Option_1 
        ON Ordered_Options.Ordered_Item_ID = Ordered_Item.ID
    JOIN (SELECT Ordered_Options.Value FROM Ordered_Options Where Option_Number = 44) as Option_2 
        ON Ordered_Options.Ordered_Item_ID = Ordered_Item.ID;

Thanks! Joe

Mysql Solutions


Solution 1 - Mysql

The easiest way would be to make use of the GROUP_CONCAT group function here..

select
  ordered_item.id as `Id`,
  ordered_item.Item_Name as `ItemName`,
  GROUP_CONCAT(Ordered_Options.Value) as `Options`
from
  ordered_item,
  ordered_options
where
  ordered_item.id=ordered_options.ordered_item_id
group by
  ordered_item.id

Which would output:

Id              ItemName       Options

1               Pizza          Pepperoni,Extra Cheese

2               Stromboli      Extra Cheese

That way you can have as many options as you want without having to modify your query.

Ah, if you see your results getting cropped, you can increase the size limit of GROUP_CONCAT like this:

SET SESSION group_concat_max_len = 8192;

Solution 2 - Mysql

I appreciate the help, I do think I have found a solution if someone would comment on the effectiveness I would appreciate it. Essentially what I did is. I realize it is somewhat static in its implementation but I does what I need it to do (forgive incorrect syntax)

SELECT
  ordered_item.id as `Id`,
  ordered_item.Item_Name as `ItemName`,
  Options1.Value
  Options2.Value
FROM ORDERED_ITEMS
LEFT JOIN (Ordered_Options as Options1)
    ON (Options1.Ordered_Item.ID = Ordered_Options.Ordered_Item_ID 
        AND Options1.Option_Number = 43)
LEFT JOIN (Ordered_Options as Options2)
    ON (Options2.Ordered_Item.ID = Ordered_Options.Ordered_Item_ID
        AND Options2.Option_Number = 44);

Solution 3 - Mysql

If you really need multiple columns in your result, and the amount of options is limited, you can even do this:

select
  ordered_item.id as `Id`,
  ordered_item.Item_Name as `ItemName`,
  if(ordered_options.id=1,Ordered_Options.Value,null) as `Option1`,
  if(ordered_options.id=2,Ordered_Options.Value,null) as `Option2`,
  if(ordered_options.id=43,Ordered_Options.Value,null) as `Option43`,
  if(ordered_options.id=44,Ordered_Options.Value,null) as `Option44`,
  GROUP_CONCAT(if(ordered_options.id not in (1,2,43,44),Ordered_Options.Value,null)) as `OtherOptions`
from
  ordered_item,
  ordered_options
where
  ordered_item.id=ordered_options.ordered_item_id
group by
  ordered_item.id

Solution 4 - Mysql

If you know you're going to have a limited number of max options then I would try this (example for max of 4 options per order):

Select OI.ID, OI.Item_Name, OO1.Value, OO2.Value, OO3.Value, OO4.Value

FROM Ordered_Items OI LEFT JOIN Ordered_Options OO1 ON OO1.Ordered_Item_ID = OI.ID LEFT JOIN Ordered_Options OO2 ON OO2.Ordered_Item_ID = OI.ID AND OO2.ID != OO1.ID LEFT JOIN Ordered_Options OO3 ON OO3.Ordered_Item_ID = OI.ID AND OO3.ID != OO1.ID AND OO3.ID != OO2.ID LEFT JOIN Ordered_Options OO4 ON OO4.Ordered_Item_ID = OI.ID AND OO4.ID != OO1.ID AND OO4.ID != OO2.ID AND OO4.ID != OO3.ID

GROUP BY OI.ID, OI.Item_Name

The group by condition gets rid of all of the duplicates that you would otherwise get. I've just implemented something similar on a site I'm working on where I knew I'd always have 1 or 2 matched in my child table, and I wanted to make sure I only had 1 row for each parent item.

Solution 5 - Mysql

What you want is called a pivot, and it's not directly supported in MySQL, check this answer out for the options you've got:

https://stackoverflow.com/questions/649802/how-to-pivot-a-mysql-entity-attribute-value-schema

Solution 6 - Mysql

Here is how you would construct your query for this type of requirement.

select ID,Item_Name,max(Flavor) as Flavor,max(Extra_Cheese) as Extra_Cheese
	from (select i.*,
					case when o.Option_Number=43 then o.value else null end as Flavor,
					case when o.Option_Number=44 then o.value else null end as Extra_Cheese
				from Ordered_Item i,Ordered_Options o) a
	group by ID,Item_Name;

You basically "case out" each column using case when, then select the max() for each of those columns using group by for each intended item.

Solution 7 - Mysql

Joe Edel's answer to himself is actually the right approach to resolve the pivot problem.

Basically the idea is to list out the columns in the base table firstly, and then any number of options.value from the joint option table. Just left join the same option table multiple times in order to get all the options.

What needs to be done by the programming language is to build this query dynamically according to a list of options needs to be queried.

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJoe EdelView Question on Stackoverflow
Solution 1 - MysqlWouter van NifterickView Answer on Stackoverflow
Solution 2 - MysqlJoe EdelView Answer on Stackoverflow
Solution 3 - MysqlWouter van NifterickView Answer on Stackoverflow
Solution 4 - MysqlSebView Answer on Stackoverflow
Solution 5 - MysqlVinko VrsalovicView Answer on Stackoverflow
Solution 6 - Mysqlck.tanView Answer on Stackoverflow
Solution 7 - MysqlhailongView Answer on Stackoverflow