I need to write code that increases the salary of an employee who is over 40 years old.
Here is my code:
DECLARE
CURSOR kurs IS SELECT ID_PRACOWNIKA , pensja_BR, wiek FROM PRACOWNICY p , OSOBY o ;
ID_PRACOWNIKA decimal(2):=0;
pensja DECIMAL(8,2);
wiek DECIMAL(2);
BEGIN
OPEN kurs;
LOOP
IF wiek > 40
THEN
UPDATE PRACOWNICY
SET pensja = PENSJA_BR * 1.02
WHERE ID_PRACOWNIKA = ID_PRACOWNIKA;
dbms_OUTPUT.put_line( ID_PRACOWNIKA|| '-'||pensja);
END IF;
ID_PRACOWNIKA := ID_PRACOWNIKA+1;
EXIT WHEN ID_PRACOWNIKA=6;
END LOOP;
CLOSE kurs;
END;
Unfortunately I have SQL error
SQL Error [6550] [65000]: ORA-06550: line 14, column 12:
PL/SQL: ORA-00904: "PENSJA": invalid identifier
ORA-06550: line 13, column 7:
PL/SQL: SQL Statement ignored
Osoby table strucuture:
Id_osoby NUMBER CONSTRAINT osoby_pk PRIMARY KEY,
Imie VARCHAR2(15) NOT NULL,
Nazwisko VARCHAR2(30) NOT NULL,
Wiek NUMBER NOT NULL CONSTRAINT ch_wiek CHECK((Wiek>=0) AND (Wiek<=125)),
Stan_cywilny VARCHAR2(12) NOT NULL,
Telefon VARCHAR2(20),
Pesel CHAR(11) NOT NULL CONSTRAINT osoba_uni UNIQUE,
Id_adresu NUMBER NOT NULL,
CONSTRAINT os_ad_fk FOREIGN KEY (Id_adresu) REFERENCES Adresy(Id_adresu)
Pracownicy table structure:
Id_pracownika NUMBER CONSTRAINT pracownik_pk PRIMARY KEY,
Id_osoby NUMBER NOT NULL CONSTRAINT pr_unique UNIQUE,
Id_stanowiska NUMBER NOT NULL,
Staz NUMBER NOT NULL CONSTRAINT ch_staz CHECK((Staz>=0) AND (Staz<=45)),
Pensja_br NUMBER NOT NULL CONSTRAINT pen_staz CHECK(Pensja_br>=1226),
CONSTRAINT pr_os_fk FOREIGN KEY (Id_osoby) REFERENCES Osoby(Id_osoby),
CONSTRAINT pr_st_fk FOREIGN KEY (Id_stanowiska) REFERENCES Stanowiska(Id_stanowiska)
I think you can use the single update statement but as you want to print the details also. You can go with
FOR
loop as follows:If you just want to update the data with a single query then you can use the following update SQL: