SELECT distinct values for multiple rows of same ID
You can use this simple solution:
SELECT DISTINCT a.id, b.value AS SIGN_UP, c.value AS FIRST_NAME, d.value AS STREET FROM tbl a LEFT JOIN tbl b ON a.id = b.id AND b.field_name = 'sign_up' LEFT JOIN tbl c ON a.id = c.id AND c.field_name = 'first_name' LEFT JOIN tbl d ON a.id = d.id AND d.field_name = 'street'
Just to be safe, I made the joins
LEFT JOIN's because I do not know if an id can have missing fields, in which case they will show up as
NULL in our derived columns.
You could also try pivoting with the help of grouping and conditional aggregating:
SELECT ID, MAX(CASE FIELD_NAME WHEN 'sign_up' THEN VALUE END) AS SIGN_UP, MAX(CASE FIELD_NAME WHEN 'first_name' THEN VALUE END) AS FIRST_NAME, MAX(CASE FIELD_NAME WHEN 'street' THEN VALUE END) AS STREET FROM atable GROUP BY ID ;
Adapted from another answer by me:
SELECT ids.ID AS ID, sign_up.VALUE AS SIGN_UP, first_name.VALUE AS FIRST_NAME, street.VALUE AS STREET FROM (SELECT DISTINCT ID FROM tableName) AS ids LEFT JOIN tableName AS sign_up ON (sign_up.ID = ids.ID AND sign_up.FIELD_NAME = 'sign_up') LEFT JOIN tableName AS first_name ON (first_name.ID = ids.ID AND first_name.FIELD_NAME = 'first_name') LEFT JOIN tableName AS street ON (street.ID = ids.ID AND street.FIELD_NAME = 'street')
The left joins will ensure that missing values will result in
NULL cells, instead of an omission of the whole row. Not sure whether that is important in your application. If it is not, you can use an inner join and in particular get rid of the subquery to select all unique IDs. See my original answer from which I derived this.
Admin 6 months
I have a table that looks like this:
ID | FIELD_NAME | VALUE 23 | sign_up | yes 23 | first_name | Fred 23 | street | Barber Lane 24 | sign_up | no 24 | first_name | Steve 24 | street | Camaro St. 25 | sign_up | yes 25 | first_name | Larry 25 | street | Huckleberry Ave
I want to run a query that will select unique ID's and the values as named columns so it would appear like so:
ID | SIGN_UP | FIRST_NAME | STREET | 23 | yes | Fred | Barber Lane | 24 | no | Steve | Camaro St. | 25 | yes | Larry | Huckleberry Ave. |
Any help would be much appreciated!!