我需要编写一个SQL查询来在SQL
中打印以下的aphanumberic序列0001, 0002, ......, 0009 000A, ......, 000Z, ......, 0010, 0011, ......, 001A, ......等等...... ZZZZ
请注意:所有字符都是大写的。
提前致谢
答案 0 :(得分:10)
你可以创建一个这样的函数:
create function to_base_36 (n integer) return varchar2
is
q integer;
r varchar2(100);
begin
q := n;
while q >= 36 loop
r := chr(mod(q,36)+case when mod(q,36) < 10 then 48 else 55 end) || r;
q := floor(q/36);
end loop;
r := chr(mod(q,36)+case when mod(q,36) < 10 then 48 else 55 end) || r;
return lpad(r,4,'0');
end;
然后像这样使用它:
select rownum, to_base_36(rownum)
from dual
connect by level < 36*36*36*36;
或者,不创建函数:
with digits as
( select n, chr(mod(n,36)+case when mod(n,36) < 10 then 48 else 55 end) d
from (Select rownum-1 as n from dual connect by level < 37)
)
select d1.n*36*36*36 + d2.n*36*36 + d3.n*36 + d4.n, d1.d||d2.d||d3.d||d4.d
from digits d1, digits d2, digits d3, digits d4
答案 1 :(得分:1)
您可以使用此功能:
create or replace FUNCTION SEQGEN(vinp in varchar2, iSeq in INTEGER)
RETURN VARCHAR2 is vResult VARCHAR2(32);
iBas INTEGER; iRem INTEGER; iQuo INTEGER; lLen CONSTANT INTEGER := 2;
BEGIN
iBas := length(vInp);
iQuo := iSeq;
WHILE iQuo > 0 LOOP
iRem := iQuo mod iBas;
--dbms_output.put_line('Now we divide ' || lpad(iQuo,lLen,'0') || ' by ' || lpad(iBas,lLen,'0') || ', yielding a quotient of ' || lpad( TRUNC(iQuo / iBas) ,lLen,'0') || ' and a remainder of ' || lpad(iRem,lLen,'0') || ' giving the char: ' || substr(vInp, iRem, 1));
iQuo := TRUNC(iQuo / iBas);
If iRem < 1 Then iRem := iBas; iQuo := iQuo - 1; End If;
vResult := substr(vInp, iRem, 1) || vResult;
END LOOP;
RETURN vResult;
END SEQGEN;
尝试功能:
SELECT * FROM (
SELECT seqgen('0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ',rownum + 47989 --start value
) Output, level evt FROM dual CONNECT BY level < 1679618) --stop value
WHERE mod(evt,50000) = 0 OR output in ('0001','0002','0009','000A','000Z',
'0010','0011','001A','ZZZZ')
请注意,如果更改字符串,则还必须更改开始和结束值。
在此处详细了解数字系统:Number System Conversion - Explanation
答案 2 :(得分:0)
-- To get 00000 to ZZZZZ next auto alphanumeric sequence using this function [Please verify before use]
-- This starts from 0-9 then A-Z and then increase next digit from 0-9 then A-Z
-- You need to pass the starting/Last sequence as value to get next sequence
CREATE OR REPLACE FUNCTION return_next_seq (curr_sequence VARCHAR2)
RETURN VARCHAR2 IS
retval VARCHAR2(4000) := NULL;
retMaxval VARCHAR2(4000) := NULL;
eval_digit CHAR(1) := NULL;
original_sequence VARCHAR2(4000) := curr_sequence;
curr1_sequence VARCHAR2(4000) := curr_sequence;
BEGIN
retval := original_sequence;
FOR j IN REVERSE 1..LENGTH(curr1_sequence) LOOP -- Using reverse to know
-- the exact digit position
eval_digit := SUBSTR(curr1_sequence, LENGTH(curr1_sequence));
--IF (ASCII(eval_digit) BETWEEN 49 AND 56) OR
--(ASCII(eval_digit) BETWEEN 97 AND 121) THEN
IF (ASCII(eval_digit) BETWEEN 48 AND 56) OR
(ASCII(eval_digit) BETWEEN 65 AND 89) THEN
eval_digit := CHR(ASCII(eval_digit) +1);
curr1_sequence := SUBSTR(curr1_sequence,1,LENGTH(curr1_sequence)-1);
retval := curr1_sequence || eval_digit || SUBSTR(original_sequence,
LENGTH(curr1_sequence || eval_digit)+1);
EXIT;
ELSE -- move to the next digit leaving the evaluated digit untouched.
IF (ASCII(eval_digit) = 57) THEN
eval_digit := CHR(ASCII(eval_digit) +8);
curr1_sequence := SUBSTR(curr1_sequence,1,LENGTH(curr1_sequence)-1);
retval := curr1_sequence || eval_digit || SUBSTR(original_sequence,
LENGTH(curr1_sequence || eval_digit)+1);
EXIT;
END IF;
IF (ASCII(eval_digit) = 90) THEN
retMaxval := eval_digit;
eval_digit := CHR(ASCII(eval_digit) -42);
curr1_sequence := SUBSTR(curr1_sequence,1,LENGTH(curr1_sequence)-1);
FOR k IN REVERSE 1..LENGTH(curr1_sequence) LOOP
IF (ASCII(SUBSTR(curr1_sequence,LENGTH(curr1_sequence))) BETWEEN 48 AND 56) OR (ASCII(SUBSTR(curr1_sequence,LENGTH(curr1_sequence))) BETWEEN 65 AND 89) THEN
retval := SUBSTR(curr1_sequence,0,LENGTH(curr1_sequence)-1) || CHR(ASCII(SUBSTR(curr1_sequence,LENGTH(curr1_sequence)))+1) || eval_digit || SUBSTR(retval,
LENGTH(curr1_sequence || eval_digit)+1);
ELSE
IF ASCII(SUBSTR(curr1_sequence,LENGTH(curr1_sequence))) = 57 THEN
retval := SUBSTR(curr1_sequence,0,LENGTH(curr1_sequence)-1) || CHR(65) || eval_digit || SUBSTR(retval,
LENGTH(curr1_sequence || eval_digit)+1);
ELSE
IF ASCII(SUBSTR(curr1_sequence,LENGTH(curr1_sequence))) = 90 AND LENGTH(curr1_sequence)>1 THEN
retval := SUBSTR(curr1_sequence,0,LENGTH(curr1_sequence)-1) || CHR(48) || eval_digit || SUBSTR(retval,
LENGTH(curr1_sequence || eval_digit)+1);
curr1_sequence := SUBSTR(curr1_sequence,1,LENGTH(curr1_sequence)-1);
retMaxval := retMaxval||'Z';
ELSE
retMaxval := retMaxval||'Z';
EXIT;
END IF;
END IF;
END IF;
END LOOP;
EXIT;
ELSE
curr1_sequence := SUBSTR(curr1_sequence,1,LENGTH(curr1_sequence)-1);
END IF;
END IF;
END LOOP;
IF retval IS NULL OR (LENGTH(retval) = LENGTH(retMaxval)) THEN
RETURN 'MAX';
END IF;
RETURN retval;
END;
-- To verify, call this function like
SELECT return_next_seq('ZY9Z') FROM dual;
-- Any improvement suggestion is welcome!
--- Thanks!..Sanjiv