我想使用Oracle SQL为用户生成唯一的sAMAccountName。生成唯一名称的逻辑是名字的首字母+整个姓氏。
如果组合已经存在，请在唯一名称的末尾附加1。 如果第二个组合已经存在，则递增直到达到唯一性。
例如，名为“ Tom Kelly”的5个用户和名为“ Tom Ke”的3个用户将具有以下唯一名称：
我已经为此编写了一个程序，但是它不能完全正常工作。我认为该查询过程中的逻辑需要更新：
SELECT SUBSTR(Custom_Unique_Name,-1,1) INTO lastletter
FROM (SELECT * FROM (SELECT Custom_Unique_Name FROM
P_USERS WHERE Custom_Unique_Name= item.tempid2 AND
user_id <> item.user_id ORDER BY Custom_Unique_Name
DESC) WHERE ROWNUM<=1);
只读表P_USERS包含所有主键为"user_id"的用户，并且唯一名称存储在属性"Custom_Unique_Name"
计算唯一名称后，需要更新主用户表"PRIMARY_UNIQUE_NAME"中的属性"T_MASTER_USERS"。该表的主键也是"user_id"，它的值与只读表"P_USERS"
DECLARE
lastletter varchar2(10);
anyexists varchar2(10);
lastfive varchar2(10);
CURSOR c_length IS SELECT SUBSTR(first_name,1,1)||(last_name) as
tempid2,LENGTH(last_name) as lengthln, user_id, Custom_Unique_Name
FROM P_USERS;
status varchar2(10);
BEGIN
FOR item in c_length
LOOP
EXIT WHEN c_length%notfound;
lastletter:=0;
SELECT NVL(Custom_Unique_Name,0) INTO status FROM P_USERS
WHERE USER_ID=item.user_id;
IF status <> '0' THEN /* Checks if the person already has a
unique id */
NULL; /* If yes, then do nothing and exit the program */
ELSE
SELECT COUNT(*) INTO anyexists FROM (SELECT
Custom_Unique_Name FROM P_USERS
WHERE Custom_Unique_Name = item.tempid2 AND user_id <>
item.user_id);
IF anyexists=0 THEN /* Check if unique id exists for
the user. If not, then assign unique_id to the user */
UPDATE T_MASTER_USERS SET
PRIMARY_UNIQUE_NAME=item.tempid2 WHERE
user_id=item.user_id;
ELSE
SELECT SUBSTR(Custom_Unique_Name,-1,1) INTO lastletter
FROM (SELECT * FROM (SELECT Custom_Unique_Name FROM
P_USERS WHERE Custom_Unique_Name= item.tempid2 AND
user_id <> item.user_id ORDER BY Custom_Unique_Name
DESC) WHERE ROWNUM<=1);
IF LENGTH(TRIM(TRANSLATE(lastletter, '
+-.0123456789',' '))) > 0 THEN /* Checks if the count
of equivalent unique names if a number. If not, append
1*/
UPDATE T_MASTER_USERS SET
PRIMARY_UNIQUE_NAME=item.tempid2||'1' WHERE
user_id=item.user_id;
ELSE
lastletter:=lastletter+1;
UPDATE T_MASTER_USERS SET
PRIMARY_UNIQUE_NAME=item.tempid2||lastletter WHERE
user_id=item.user_id;
END IF;
END IF;
END IF;
END LOOP;
END;
答案 0 :(得分：1)
这样的事情怎么样？
SQL> create table users
2 (username varchar2(20) primary key);
Table created.
SQL> create or replace procedure p_un (par_first_name in varchar2,
2 par_last_name in varchar2)
3 is
4 l_username users.username%type;
5 l_letter users.username%type;
6 l_digit users.username%type;
7 retval users.username%type;
8 begin
9 l_username := upper(substr(par_first_name, 1, 1) || par_last_name);
10
11 select max(regexp_substr(u.username, '^\w+')) l_letter,
12 max(regexp_substr(u.username, '\d+$')) l_digit
13 into l_letter,
14 l_digit
15 from users u
16 where u.username like l_username ||'%'
17 and ( substr(u.username, length(l_username) + 1, 1) between '1' and '9'
18 or ( regexp_substr(u.username, '\d$') is null
19 and substr(u.username, length(l_username) + 1, 1) is null
20 )
21 )
22 and ( to_number(regexp_substr(username, '\d$')) =
23 (select max(to_number(regexp_substr(u1.username, '\d$')))
24 from users u1
25 where u1.username like l_username ||'%'
26 and ( substr(u1.username, length(l_username) + 1, 1) between '1' and '9'
27 or ( regexp_substr(u1.username, '\d$') is null
28 and substr(u1.username, length(l_username) + 1, 1) is null
29 )
30 )
31 )
32 or regexp_substr(u.username, '\d$') is null
33 );
34
35 if l_letter is null then
36 retval := upper(l_username);
37 else
38 retval := upper(l_username || to_char(to_number(nvl(l_digit, 0)) + 1));
39 end if;
40
41 dbms_output.put_line('l_username = ' || l_username||', letter = '|| l_letter||
42 ', digit = '||l_digit ||', new username = ' || retval);
43 insert into users (username) values (retval);
44 end;
45 /
Procedure created.
它是做什么的？
SELECT（第11行）搜索现有的用户名，
1添加到最大数字并将其附加到默认用户名值（第38行）测试：
SQL> exec p_un('Tom', 'Kelly');
l_username = TKELLY, letter = , digit = , new username = TKELLY
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Kelly');
l_username = TKELLY, letter = TKELLY, digit = , new username = TKELLY1
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Kelly');
l_username = TKELLY, letter = TKELLY1, digit = 1, new username = TKELLY2
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Ke');
l_username = TKE, letter = , digit = , new username = TKE
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Ke');
l_username = TKE, letter = TKE, digit = , new username = TKE1
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Kelly');
l_username = TKELLY, letter = TKELLY2, digit = 2, new username = TKELLY3
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Ke');
l_username = TKE, letter = TKE1, digit = 1, new username = TKE2
PL/SQL procedure successfully completed.
SQL> exec p_un('Tom', 'Ke');
l_username = TKE, letter = TKE2, digit = 2, new username = TKE3
PL/SQL procedure successfully completed.
结果：
SQL> select * From users;
USERNAME
--------------------
TKE
TKE1
TKE2
TKE3
TKELLY
TKELLY1
TKELLY2
TKELLY3
8 rows selected.
答案 1 :(得分：0)
为您的情况尝试以下“匿名”阻止：
create table P_USERS
(first_name varchar2(20),
last_name varchar2(20),
user_id number);
insert into P_USERS values ('Tom','Kelly',1);
insert into P_USERS values ('Tom','Kelly',2);
insert into P_USERS values ('Tom','Kelly',3);
insert into P_USERS values ('Tom','Kelly',4);
insert into P_USERS values ('Tom','Kelly',5);
insert into P_USERS values ('Tom','Ke',6);
insert into P_USERS values ('Tom','Ke',7);
insert into P_USERS values ('Tom','Ke',8);
commit;
create table T_MASTER_USERS
( user_id number,
PRIMARY_UNIQUE_NAME varchar2(20));
insert into T_MASTER_USERS values (1,NULL);
insert into T_MASTER_USERS values (2,NULL);
insert into T_MASTER_USERS values (3,NULL);
insert into T_MASTER_USERS values (4,NULL);
insert into T_MASTER_USERS values (5,NULL);
insert into T_MASTER_USERS values (6,NULL);
insert into T_MASTER_USERS values (7,NULL);
insert into T_MASTER_USERS values (8,NULL);
commit;
DECLARE
cursor uniq_grp is SELECT SUBSTR(first_name,1,1)||(last_name) as
tempid2 ,count(*) as CNT
FROM P_USERS
group by SUBSTR(first_name,1,1)||(last_name);
cursor concat_names is SELECT SUBSTR(first_name,1,1)||(last_name) as
tempid2, user_id
FROM P_USERS;
V_CNT number;
BEGIN
FOR C1 in uniq_grp
LOOP
FOR C2 IN concat_names
LOOP
IF C2.tempid2=C1.tempid2 THEN
V_CNT := c1.CNT -1;
UPDATE T_MASTER_USERS SET
PRIMARY_UNIQUE_NAME=c2.tempid2||V_CNT WHERE
user_id=C2.user_id;
c1.cnt := V_CNT;
END IF;
END LOOP;
COMMIT;
END LOOP;
END;
答案 2 :(得分：0)
创建或替换过程generate_unique（varchar2中的par_first_name，varchar2中的par_last_name）
（varchar2中的par_first_name，varchar2中的par_last_name）
是
l_username users.username％type;
l_letter users.username％type;
开始
l_username := upper(substr(par_first_name, 1, 1) || par_last_name);
dbms_output.put_line('l_username = ' || l_username );
当MAX（REGEXP_REPLACE（UPPER（USERNAME），'[^ 0-9]'，''））为NULL时选择情况
THEN l_username
ELSE l_username
||
(MAX(REGEXP_REPLACE(UPPER(USERNAME),'[^0-9]', ''))+1)
END into l_letter FROM USERS WHERE REGEXP_LIKE(UPPER(USERNAME),l_username||'[0-9]*$');
dbms_output.put_line（'唯一的用户ID为'|| l_letter）;
结束;