Why Oracle is saying not a GROUP BY expression?

Question

I am Trying to retrive data from the Oracle database table with following conditions

1. All entries found by zip-code should be grouped, sorted by zip-code desce

Answer 1

The specific answer to your question is that uuid is not included in the group by expression. Other answers have already explained the MySQL (mis) feature that allows this to happen. Fixing the query is easy:

SELECT ba.uuid AS uuid, COUNT(*) over() AS rowcount
FROM basicaddress ba 
WHERE ba.postalcode='143456' OR ba.lastname LIKE '%143456%' OR ba.city LIKE '%143456%'
GROUP BY uuid,
        CASE WHEN ba.postalcode='143456' THEN ba.postalcode END 
ORDER BY 
        max(CASE WHEN ba.postalcode='143456' THEN ba.postalcode END) DESC, 
        max(CASE WHEN ba.lastname LIKE '%143456%' THEN ba.lastname END) ASC, 
        max(CASE WHEN ba.city LIKE '%143456%' THEN ba.city ELSE ba.postalcode END)

I also removed the "distinct" which should be redundant when using group by.

That said, I don't see how this satisfies your original question. I would expect something more along the lines of the following:

select (case when postalcode is not null then postalcode
             when city is not null then city
             when lastname is not null then lastname
        end),
       (case when postalcode is not null then 1
             when city is not null then 2
             when lastname is not null then 3
        end),
       count(*)
from basicaddress ba
group by (case when postalcode is not null then postalcode
               when city is not null then city
               when lastname is not null then lastname
          end),
         (case when postalcode is not null then 1
             when city is not null then 2
             when lastname is not null then 3
        end)
order by 1, 2

This may not be exactly what you need, but it should point you in the right direction. Also, this is standard SQL and will work in both databases.

Answer 2

This happens to you just because MySQL breaks the logic of SQL.

Let's say we have table emp:

id ename dept
1  mark  10
2  John  10
3  Mary  10
4  Jane  20

and the query:

select dept, ename
from emp 
group by dept;

you'll get what? You should get two lines, because there are two departments, but the query ask for ename. For 20 is clear but for 10 the engine should return what?

It shoud return an error. Can't guess what ename to give. Oracle shoot an error - your error, but MySQL gets an ename(not guaranteed which). That's missleading and may conduct bugs.

Correct queries would be:

select dept, max(ename) --the latest, alaphabeticaly
from emp 
group by dept;

and

--all enames and groups
select dept, ename 
from emp 
group by dept, ename;

After you correct this part, you'll have to resolve the

COUNT(*) over() AS rowcount

part. In oracle, AFAIK, you can't mix analytic functions with group by queries.

Answer 3

I don't know about MySQL, but following works in Oracle 11g:

With 
T_PATTERN as (select '%143456%' as PATTERN from DUAL)
, T_INPUT as (
  select 1 as UUID, 'X143456y' as FIRSTNAME, 'last1' as LASTNAME, 'u143456v' as CITY, 'w143456z' as POSTALCODE from DUAL union
  select 2 as UUID, 'first2' as FIRSTNAME, 'a143456b' as LASTNAME, 'a143456b' as CITY, 'postal2' as POSTALCODE from DUAL union
  select 3 as UUID, 'first3' as FIRSTNAME, 'last3' as LASTNAME, 'c143456d' as CITY, 'postal3' as POSTALCODE from DUAL union
  select 4 as UUID, 'first4' as FIRSTNAME, 'last4' as LASTNAME, 'city4' as CITY, 'postal4' as POSTALCODE from DUAL
),
Q_FNAME as (
  select TI.UUID, TI.FIRSTNAME, COUNT(*) FN_COUNT from T_INPUT TI inner join T_PATTERN TP on 1=1 
  where UPPER(TI.FIRSTNAME) like TP.PATTERN
  GROUP BY TI.UUID, TI.FIRSTNAME
),
Q_LNAME as (
  select TI.UUID, TI.LASTNAME, COUNT(*) LN_COUNT from T_INPUT TI inner join T_PATTERN TP on 1=1 
  where UPPER(TI.LASTNAME) like TP.PATTERN
  group by TI.UUID, TI.LASTNAME
),
Q_CITY as (
  select TI.UUID, TI.CITY, COUNT(*) CT_COUNT from T_INPUT TI inner join T_PATTERN TP on 1=1 
  where UPPER(TI.CITY) like TP.PATTERN
  group by TI.UUID, TI.CITY
),
Q_PCODE as (
  select TI.UUID, TI.POSTALCODE, COUNT(*) PC_COUNT from T_INPUT TI inner join T_PATTERN TP on 1=1 
  where UPPER(TI.POSTALCODE) like TP.PATTERN
  group by TI.UUID, TI.POSTALCODE
)
select TI.UUID, TI.FIRSTNAME, TI.LASTNAME, TI.CITY, TI.POSTALCODE, 
  QF.FN_COUNT, QL.LN_COUNT, QC.CT_COUNT, QP.PC_COUNT
from T_INPUT ti
left join Q_FNAME QF on TI.UUID=QF.UUID
left join Q_LNAME QL on TI.UUID=QL.UUID
left join Q_CITY QC on TI.UUID=QC.UUID
left join Q_PCODE QP on TI.UUID=QP.UUID
order by ti.UUID;

Answer 4

It works in MySQl because MySQL has a shoddy interpretation of SQL.

In any proper dialect, anything in the SELECT or ORDER BY clauses needs to be either an aggregate function, or a GROUPed expression.

At the very least, you're going to need to group by ba.uuid, lastname and city.