MySQL query to gather incorrectly stored data - mysql

Question

I have recently taken over a email campaign project and need to generate a report for the customer. However the data has been stored very strangely.
Basically the client wants a report of the subscribers first name and last name that have subscribed to a emailing list.
Example table data.
------------------------------------------------------------
id | owner_id | list_id | field_id | email_address | value
------------------------------------------------------------
1 10 1 137 me#example.com John
2 10 1 138 me#example.com Doe
So as you can see, John Doe has subscribed to mailing list 1, and field_id 137 is his first name and field_id 138 is his last name.
The client is looking for a export with the users first name and last name all is one field.
I tred the following sql query
SELECT value
FROM Table_A AS child
INNER JOIN Table_A AS parent
ON parent.email_address = child.email_address
WHERE child.owner_id = '10'
But unfortunately the query gives me the results in many rows but not appending the first name and last name into one field,
If anyone can provide some assistance that would be awesome.
Thanks.

Answer 1

SELECT
concat( parent.value,' ',child.value)name
FROM mytable AS child
left JOIN mytable AS parent
ON parent.email_address = child.email_address
WHERE child.owner_id = '10'
and parent.field_id=137 and child.field_id=138
Check at-http://sqlfiddle.com/#!9/199b4b/45

Answer 2

I think you have to use a variable to put in there everything you have to and then select the variable with the desired name of yours.
For example:
DECLARE #yourvariable VARCHAR(MAX)
SELECT #yourvariable = COALESCE(#yourvariable + " ") + value
FROM table_A
WHERE owner_id = 10
SELECT #yourvariable as FullName
Try that, it might help.

Answer 3

You can try this code(column name equals value in your original DB):
select a.name
from
table_a a inner join table_a b
on a.email_address = b.email_address and a.field_id <> b.field_id
where a.owner_id=10
order by a.field_id
Here is the example link:
http://sqlfiddle.com/#!9/5fbdf6/25/0

Answer 4

As per assumptions, first name has the field id 137 and last name has the field id 138.
You can try the following query to get the desired result.
SELECT CONCAT(SUBSTRING_INDEX(GROUP_CONCAT(`value`),",",1)," ",SUBSTRING_INDEX(GROUP_CONCAT(`value`),",",-1)) AS client_name
FROM Table_A
WHERE owner_id = 10
AND field_id IN (137, 138)
GROUP BY email_address;