MySQL query works in phpMyAdmin but not at site. Unknown column 'zone' in 'where clause'

Question

The following query works in phpMyAdmin but doesn't work when I run it in through the website in PHP.

SELECT * FROM 
      (SELECT name, zone FROM staff 
        LEFT OUTER JOIN zones 
          ON staff.suburb=zones.suburb
      ) A WHERE zone='2'

The following query also doesn't work on the website but works in phpMyAdmin:

SELECT name, zone FROM staff 
  LEFT OUTER JOIN zones 
    ON staff.suburb=zones.suburb 
  WHERE zone='2'

Both give an error:

Unknown column 'zone' in 'where clause'.

What am I doing wrong here?

Answer 1

If you want to run the first query the way you provided, it should be more like this:

SELECT * 
FROM (
    SELECT name,zone 
    FROM staff 
    LEFT OUTER JOIN zones ON staff.suburb=zones.suburb
) A 
WHERE A.zone='2'

But I totaly do not understand why you are doing a SUB-SELECT, that is totally unnecessary. The second query should work, unless you have made a typographical error. Also it is good practice to escape every column, or table name used in query, like so:

SELECT `name`, `zone` 
FROM `staff` s 
LEFT OUTER JOIN `zones` z 
    ON s.`suburb` = z.`suburb` 
WHERE z.`zone` = '2'

Answer 2

Try this :

SELECT `name`, `zone`
FROM `staff` AS s
LEFT OUTER JOIN `zones` AS z ON `s`.`suburb`=`z`.`suburb`
WHERE `s`.`zone`='2';

Supposing that the column zone is in the table staff . If it's in the table zones , change the WHERE to :

WHERE `z`.`zone`='2';

The key part here is the precision in the WHERE clause. Maybe MySQL don't know where to look up (in which table), so adding a precision on the table will reduce the risk. Moreover, make sur this column exists in that table (and with that typo. zones or Zone could lead to errors).