
Learn the cause and how to resolve the ORA-06550 error message in Oracle.
Description
When you encounter an ORA-06550 error, the following error message will appear:
- ORA-06550: line num, column num: str
Cause
You tried to execute an invalid block of PLSQL code (like a stored procedure or function), but a compilation error occurred.
Resolution
The option(s) to resolve this Oracle error are:
Option #1
Refer to the line and column numbers (in the error message) to find the compilation error and correct it. Then try recompiling your code.
Let’s look at an example of how to resolve an ORA-06550 error. For example, if you created a procedure called TestProc as follows:
SQL> CREATE OR REPLACE PROCEDURE TestProc 2 AS 3 vnum number; 4 BEGIN 5 vnum := vAnotherNum; 6 END; 7 / Warning: Procedure created with compilation errors.
This procedure was created with compilation errors. So if we try to execute this procedure, we will get an ORA-06550 error as follows:
SQL> execute TestProc(); BEGIN TestProc(); END; * ERROR at line 1: ORA-06550: line 1, column 7: PLS-00905: object EXAMPLE.TESTPROC is invalid ORA-06550: line 1, column 7: PL/SQL: Statement ignored
You can run the SHOW ERROR command to view the errors as follows:
SQL> show error procedure TestProc; Errors for PROCEDURE TESTPROC: LINE/COL ERROR -------- ----------------------------------------------------------------- 5/1 PL/SQL: Statement ignored 5/9 PLS-00201: identifier 'VANOTHERNUM' must be declared
As you can see, the error is caused by the variable called VANOTHERNUM not being declared. To resolve this error, we can modify our TestProc procedure to declare the variable as follows:
SQL> CREATE OR REPLACE PROCEDURE TestProc 2 AS 3 vnum number; 4 vAnotherNumber number; 5 BEGIN 6 vAnotherNum := 999; 7 vnum := vAnotherNum; 8 END; 9 / Procedure created.
And now when we execute our TestProc procedure, the ORA-06550 error has been resolved.
SQL> execute TestProc(); PL/SQL procedure successfully completed.
Этот код SQL выдает мне ошибку «Ошибка в строке 2: PL / SQL: инструкция проигнорирована». Я работаю над приложением SQL oracle express / APEX: я пробовал все, что мог придумать, и каждый раз это вызывает у меня разные проблемы.
CREATE or replace TRIGGER remove_artista
instead of delete on V_ARTISTA
REFERENCING old AS orow
FOR EACH ROW
BEGIN
if exists(select * from Utilizadores where pessoaID = orow.pessoaID) then
delete from Pessoas where pessoaID = orow.pessoaID;
ELSE
delete from Artistas where pessoaID = orow.pessoaID;
delete from Pessoas where pessoaID = orow.pessoaID;
end if;
END;
Вид:
create or replace view v_artista as
select
pessoaID, nome_p, sexo, data_nasc, nome_art, biografica
from Pessoas natural inner join Artistas;
РЕДАКТИРОВАТЬ: исправлена небольшая опечатка в коде.
2 ответа
Лучший ответ
Полная ошибка, которую я получаю от вашего триггера, выглядит следующим образом:
LINE/COL ERROR
-------- -----------------------------------------------------------------
2/1 PL/SQL: Statement ignored
2/4 PLS-00204: function or pseudo-column 'EXISTS' may be used inside
a SQL statement only
По сути, проблема в том, что вы не можете сказать if exists(...), как вы это делаете. Oracle не позволяет вам.
Вместо этого попробуйте выбрать количество совпадающих строк в таблице Utilizadores в локальной переменной, а затем использовать это в своем операторе if:
CREATE or replace TRIGGER remove_artista
instead of delete on V_ARTISTA
REFERENCING old AS orow
FOR EACH ROW
DECLARE
l_count INTEGER;
BEGIN
select count(*)
into l_count
from Utilizadores
where pessoaID = :orow.pessoaID;
if l_count > 0 then
delete from Pessoas where pessoaID = :orow.pessoaID;
ELSE
delete from Artistas where pessoaID = :orow.pessoaID;
delete from Pessoas where pessoaID = :orow.pessoaID;
end if;
END;
Мне также нужно было заменить orow на :orow. После внесения этого изменения ваш триггер скомпилирован для меня.
4
Luke Woodward
24 Май 2014 в 00:06
Я не думаю, что вы можете использовать конструкцию IF EXISTS, чтобы проверить, существует ли строка. Вы можете использовать SELECT COUNT(*) INTO <a variable>. Однако вам может не понадобиться проверять, существует ли строка. Следующий код, вероятно, подойдет:
CREATE OR REPLACE TRIGGER remove_artista
INSTEAD OF DELETE ON V_ARTISTA
FOR EACH ROW
BEGIN
DELETE FROM PESSOAS
WHERE PESSOAID = :OLD.PESSOAID;
DELETE FROM Artistas
WHERE PESSOAID = :OLD.PESSOAID
AND NOT EXISTS (SELECT 1 FROM UTILIZADORES WHERE PESSOAID = :OLD.PESSOAID);
END;
Строка из PESSOAS в любом случае будет удалена. Строка из ARTISTAS будет удалена, только если PESSOAID не существует в UTILIZADORES.
Ссылки :
Проверить, существует ли запись на форуме OTN
2
Joseph B
24 Май 2014 в 00:11
SummaryAfter |
Applies toBizagi |
SymptomsAfter ORA-06550: line 1, column 7: The
|
CauseThe |
Solution1. grant execute on In 2. 3. |
Yesterday, while learning oracle stored procedures, I wrote a demo of stored procedures. The statement is as follows:
CREATE OR REPLACE PROCEDURE RAISESALARY(PNAME IN VARCHAR2(20)) AS psssal TESTDELETE.TESTID%TYPE; BEGIN SELECT TESTID INTO psssal FROM TESTDELETE WHERE TESTNAME=PNAME; UPDATE TESTDELETE SET TESTID=(TESTID+10000) WHERE TESTNAME=PNAME; DBMS_OUTPUT.PUT_LINE('The original salary'||psssal||' After the raise'||(psssal+1000)); end; /
The idea is to find the number type field through the varchar(20) type field of the table, and then change the number type field. The table structure is as follows:
create table TESTDELETE ( TESTID NUMBER, TESTNAME VARCHAR2(20) )
Call the stored procedure, and the result is as follows:
Connected to: Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production With the Partitioning, OLAP, Data Mining and Real Application Testing options CREATE OR REPLACE PROCEDURE RAISESALARY(PNAME IN VARCHAR2(20)) 2 AS 3 psssal TESTDELETE.TESTID%TYPE; 4 BEGIN 5 SELECT TESTID INTO psssal FROM TESTDELETE WHERE TESTNAME=PNAME; UPDATE TESTDELETE SET TESTID=(TESTID+10000) WHERE TESTNAME=PNAME; DBMS_OUTPUT.PUT_LINE('The original salary'||psssal||' After the raise'||(psssal+1000)); 8 end; 9 / Warning: Procedure created with compilation errors. SET SERVEROUTPUT ON; BEGIN RAISESALARY('name2091'); COMMIT; END; SQL> 2 3 4 5 / RAISESALARY('name2091'); * ERROR at line 2: ORA-06550: line 2, column 5: PLS-00905: object TEST.RAISESALARY is invalid ORA-06550: line 2, column 5: PL/SQL: Statement ignored SQL>
An error occurred: the stored procedure is not valid.
what? The stored procedure that Mingming just established. Then I turn to the above sentence after creation, Warning: Procedure created with compilation errors.
That is to say, there is an error in creating the stored procedure. OK, look for the stored procedure to see what is wrong.
emmmm, looking back for a long time, just a few lines, I can’t see what’s wrong, baidu, google.
Later, I saw this in StackOverFlow. The original link is as follows:
https://stackoverflow.com/questions/48497140/oracle-sql-stored-procedure-object-invalid
The landlord found a saying in it:
You can’t give a size or precision restriction for the data type of a formal parameter to a function or procedure, so NUMBER(10,0) should just be NUMBER;
That is, you cannot specify the size or precision of the data for the parameters of functions and stored procedures. OK, looking back at my parameter, I see that varchar2(20) clearly specifies the precision for the parameter type. You need to change to varchar2.
Or write the table name. Field% TYPE directly.
After the change, run as follows:
CREATE OR REPLACE PROCEDURE RAISESALARY(PNAME IN VARCHAR) 2 AS 3 psssal TESTDELETE.TESTID%TYPE; 4 BEGIN 5 SELECT TESTID INTO psssal FROM TESTDELETE WHERE TESTNAME=PNAME; UPDATE TESTDELETE SET TESTID=(TESTID+10000) WHERE TESTNAME=PNAME; DBMS_OUTPUT.PUT_LINE('The original salary'||psssal||' After the raise'||(psssal+1000)); 8 end; 9 / Procedure created. BEGIN RAISESALARY('name2091'); 3 COMMIT; 4 END; 5 / The original salary2091 After the raise3091 PL/SQL procedure successfully completed. SQL>
After the modification, it is clear that there is no warning. The Procedure created appears
Run successfully! Problem solving.