[英]Control structures (IF/ELSE)
我嘗試創建plsql塊,但是它不起作用。 甲骨文說我在第2行有錯誤。我認為可能是因為替換變量放置不正確!
DECLARE
V_ENAME EMPLOYEES.LAST_NAME%TYPE := '&LNAME';
V_SAL EMPLOYEES.SALARY%TYPE;
BEGIN
SELECT LAST_NAME, SALARY
INTO V_ENAME, V_SAL
FROM employees WHERE LAST_NAME = V_ENAME;
IF V_SAL < 3000 THEN
v_sal := v_sal + 500;
DBMS_OUTPUT.PUT_LINE (v_ename || 'have increasement ');
ELSIF V_SAL > 3000 THEN
DBMS_OUTPUT.PUT_LINE (v_ename || 'do not have increasement ');
ELSE
DBMS_OUTPUT.PUT_LINE ('error');
END IF;
END;
由於可能有很多員工使用用戶傳遞的姓氏,因此可以使用隱式CURSOR FOR LOOP
。 另外,為清楚起見,在輸出中添加first_name
。
DECLARE
v_ename employees.last_name%TYPE := '&LNAME';
v_sal employees.salary%TYPE;
BEGIN
for rec IN ( SELECT first_name,last_name,salary
FROM employees WHERE last_name = v_ename
)
LOOP
IF rec.salary < 3000 THEN
v_sal := rec.salary + 500;
dbms_output.put_line (rec.first_name||' '||rec.last_name ||
' has an increase');
ELSIF rec.salary > 3000 THEN
dbms_output.put_line (rec.first_name||' '||rec.last_name ||
' does not have an increase ');
ELSE dbms_output.put_line ('error');
END IF;
END LOOP;
END;
/
執行
..
old:DECLARE
v_ename employees.last_name%TYPE := '&LNAME';
..
new:DECLARE
v_ename employees.last_name%TYPE := 'Grant';
..
Douglas Grant has an increase
Kimberely Grant does not have an increase
PL/SQL procedure successfully completed.
不知道這是否是您的問題,但是當您已經確保where子句中的last_name等於v_ename時,實際上無需在v_ename中選擇last_name。
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