如何存储选择结果到Oracle过程中的变量
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20-09-2019 - |
题
我写了一个简单的过程。我尝试将选择结果存储在变量中。我使用“选择”查询,但我无法这样做。
例子:
DECLARE
v_employeeRecord employee%ROWTYPE;
BEGIN
SELECT * INTO v_employeeRecord
FROM Employee WHERE Salary > 10;
END;
解决方案
您有几个选择。您可以将该查询变成光标:
DECLARE
CURSOR v_employeeRecords IS
SELECT * FROM Employee WHERE Salary > 10;
v_employeeRecord employee%ROWTYPE;
BEGIN
FOR v_employeeRecord IN v_employeeRecords LOOP
/* Do something with v_employeeRecord */
END LOOP;
END;
或者,您可以创建一个 TABLE
多变的:
DECLARE
v_employeeRecord employee%ROWTYPE;
v_employeeRecords IS TABLE OF employee%ROWTYPE;
i BINARY_INTEGER;
BEGIN
SELECT * BULK COLLECT INTO v_employeeRecords
FROM Employee WHERE Salary > 10;
i := v_employeeRecords.FIRST;
WHILE v_employeeRecords.EXISTS(i) LOOP
v_employeeRecord := v_employeeRecords(i);
/* Do something with v_employeeRecord */
i := v_employeeRecords.NEXT(i);
END;
END;
我没有在Oracle中尝试过这些样本,因此您可能会出现编译器错误...
其他提示
如果您的选择返回多个一行,则您将无法将Select使用为Synthax。
您将需要构建一个循环以浏览结果集:
亚当 展示了您将如何使用显式光标和散装收集环。我将展示如何构建最简单的循环(隐式光标,不需要声明部分):
BEGIN
FOR c_emp IN (SELECT *
FROM Employee
WHERE Salary > 10) LOOP
/* do something with each row, for example:*/
UPDATE foo SET bar = bar + c_emp.salary WHERE id = c_emp.id;
END LOOP;
END;
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