oracle 如何为 PL/SQL 块的表达式中的变量赋值？
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how to assign value to variable in expression for a PL/SQL block?
提问by Mahesh Meniya
When I run this script it returns error in this statement no1:=(no1+no2)-(no2:=no1);
当我运行此脚本时，它在此语句中返回错误 no1:=(no1+no2)-(no2:=no1);
declare
no1 number(3):=31;
no2 number(3):=34;
begin
dbms_output.put_line('Before swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
-- no1:=(no1+no2)-(no2:=no1); generate error
dbms_output.put_line('After swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
end;
回答by Justin Cave
In addition to using the xor trick in PL/SQL, you can simply use a SQL statement
除了在 PL/SQL 中使用 xor 技巧之外，您还可以简单地使用 SQL 语句
DECLARE
a number := 17;
b number := 42;
BEGIN
SELECT a, b
INTO b, a
FROM dual;
dbms_output.put_line( 'a = ' || a );
dbms_output.put_line( 'b = ' || b );
END;
which swaps the two variables without using a temp variable
在不使用临时变量的情况下交换两个变量
SQL> ed
Wrote file afiedt.buf
1 DECLARE
2 a number := 17;
3 b number := 42;
4 BEGIN
5 SELECT a, b
6 INTO b, a
7 FROM dual;
8 dbms_output.put_line( 'a = ' || a );
9 dbms_output.put_line( 'b = ' || b );
10* END;
SQL> /
a = 42
b = 17
PL/SQL procedure successfully completed.
回答by A.B.Cade
Actually, you can also swap two numbers with no temp number, by using the Swap XOR Algorithm(but you'll still have 3 commands):
实际上，您还可以通过使用交换异或算法（但您仍然有 3 个命令）来交换两个没有临时编号的数字：
declare
no1 number(3):=31;
no2 number(3):=34;
begin
dbms_output.put_line('Before swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
n1 := (n1 + n2) - bitand(n1,n2) * 2;
n2 := (n2 + n1) - bitand(n2,n1) * 2;
n1 := (n1 + n2) - bitand(n1,n2) * 2;
dbms_output.put_line('After swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
end;
As to how do bitwise xor in plsql see here
至于如何在 plsql 中按位异或看这里
IMHO, one should avoid one-liners in real programs, it's fun to write them but hell to maintain them...
恕我直言，在实际程序中应该避免单行，编写它们很有趣，但维护它们却是地狱......
回答by denied
You can declare an additional procedure with in-out parameters like this :
您可以声明一个带有输入输出参数的附加过程，如下所示：
PROCEDURE swap(a IN OUT NUMBER, b IN OUT NUMBER) is
buff NUMBER;
BEGIN
buff := a;
a := b;
b := buff;
END swap;
and use it like this :
并像这样使用它：
swap(a, b);
回答by Korhan Ozturk
You cannotmake multiple assignment operations in a single statement, so that will keep generating errors. Instead, I suggest you to define a temp variable and use it for your swap operation, like the following:
您不能在单个语句中进行多个赋值操作，因此会不断产生错误。相反，我建议您定义一个临时变量并将其用于交换操作，如下所示：
declare
no1 number(3):=31;
no2 number(3):=34;
temp number;
begin
dbms_output.put_line('Before swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
-- no1:=(no1+no2)-(no2:=no1); generate error
temp := no1;
no1 := no2;
no2 : temp;
dbms_output.put_line('After swap');
dbms_output.put_line('No1 : '||no1||' No2 : '||no2 );
end;
