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(转发)PostgreSQL 与 Oracle 相异点

    博客分类:
  • SQL
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前段时间做了一个数据库移植的项目,对Oracle SQL 转化为PostgreSQL 有了一点了解。现将项目中遇到的一些相异点,一一分享,共同提高,共同进步。

ORACLE 与 PostgreSQL 相异点
NO 问题点 Oracle PostgreSQL
1 DUAL SELECT 1+1 FROM DUAL SELECT  1+1
或者
CREATE VIEW dual AS
      SELECT 'X'::VARCHAR(1) AS DUMMY
再 SELECT 1+1 FROM DUAL
2 NEXTVAL SELECT A_TABLE_SEQUENCE.NEXTVAL
FROM   DUAL
SELECT NEXTVAL('A_TABLE_SEQUENCE')
FROM   DUAL
3 ROWNUM ①SELECT *
FROM  AGE_TYPE
WHERE ROWNUM<=5
①SELECT *
FROM AGE_TYPE
LIMIT 5 OFFSET 0
②SELECT *
FROM AGE_TYPE
WHERE CODE IS NOT NULL
AND ROWNUM<=5
ORDER BY CODE DESC
②SELECT  *
FROM
AGE_TYPE
WHERE CODE IS NOT NULL
ORDER BY CODE DESC
LIMIT 5 OFFSET 0
4 (+) ①SELECT *
FROM A_TABLE A , B_TABLE B
WHERE A.ID(+)=B.ID
①SELECT *
FROM A_TABLE A
RIGHT OUTER JOIN
B_TABLE B
ON A.ID=B.ID
②SELECT *
FROM A_TABLE A , B_TABLE B
WHERE A.ID(+)=B.ID
AND A.COL1='COL1_VALUE'
②SELECT *
FROM A_TABLE A
RIGHT OUTER JOIN B_TABLE B
ON A.ID=B.ID AND A.COL1='COL1_VALUE'
③SELECT *
FROM A_TABLE A, B_TABLE B,C_TABLE C,D_TABLE D
WHERE
A.ID=B.ID(+) AND
A.ID=C.ID(+) AND
A.COL1=D.COL1
③SELECT *
FROM (A_TABLE A
LEFT OUTER JOIN B_TABLE B
ON A.ID=B.ID)
LEFT OUTER JOIN C_TABLE C
ON A.ID=C.ID,D_TABLE D
WHERE A.COL1=D.COL1
④!!!
SELECT *
FROM A_TABLE A
WHERE A.COL1(+)=0 AND
  A.COL2(+) ='A_VALUE2'
④!!!
SELECT *
FROM A_TABLE A
WHERE A.COL1=0 AND
  A.COL2='A_VALUE2'
WHERE (A.COL1=0 OR A.COL1 IS NULL) AND
  (A.COL2='A_VALUE2' OR A.COL2 IS NULL)
5 AS SELECT A.COL1  A_COL1,
           A.COL2  A_COL2
FROM A_TABLE A
SELECT A.COL1 AS A_COL1,
           A.COL2 AS A_COL2
FROM A_TABLE A
6 NVL SELECT NVL(SUM(VALUE11),0) FS_VALUE1,
            NVL(SUM(VALUE21),0) FS_VALUE2
FROM   FIELD_SUM 
SELECT COALESCE(SUM(VALUE11),0) AS FS_VALUE1,
           COALESCE(SUM(VALUE21),0) AS FS_VALUE2
FROM   FIELD_SUM 
7 TO_
NUMBER
SELECT COL1
FROM A_TABLE
ORDER BY TO_NUMBER(COL1)
SELECT COL1
FROM A_TABLE
ORDER BY TO_NUMBER(COL1,999999)
[注:'999999' ---- 6位数
COL1字段的 度]
8 DECODE SELECT DECODE(ENDFLAG,'1','A','B') ENDFLAG
FROM  TEST
SELECT
(CASE ENDFLAG
WHEN '1' THEN 'A'
ELSE '
B ' END) AS ENDFLAG
FROM TEST
9 时间
问题
UPDATE A_TABLE
SET ENTREDATE= SYSDATE
UPDATE A_TABLE
SET ENTREDATE=TO_TIMESTAMP(CURRENT_TIMESTAMP,'YYYY-MM-DD HH24:MI:SS')
或者
UPDATE A_TABLE
SET ENTREDATE=CURRENT_TIMESTAMP
SELECT TO_DATE('20010203','YYYY-MM-DD') AS DAY
FROM DUAL
SELECT TO_DATE('20010203','YYYYMMDD') AS DAY
FROM DUAL

SELECT TO_DATE('20010203','YYYY-MM-DD') AS DAY
FROM DUAL
SELECT TO_DATE(SYSDATE ,'YYYY-MM-DD') AS DAY
FROM DUAL
SELECT TO_DATE(CURRENT_DATE,'YYYY-MM-DD') AS DAY
FROM DUAL
SELECT TO_DATE(SYSDATE ,'YYYY/MM/DD') AS DAY
FROM DUAL
SELECT TO_DATE(CURRENT_DATE,'YYYY/MM/DD') AS DAY
FROM DUAL
10 || SELECT NULL||'-'||NULL AS VALUES1
FROM DUAL
SELECT COALESCE(NULL,'')||'-'||COALESCE(NULL,'') AS VALUES1
FROM DUAL
SELECT NULL||'-' ||NULL AS VALUES1
FROM DUAL
11 aggregate SELECT ROUND(AVG(SUM(BASICCNT1))) BASICCNT
FROM   ACCESS_INFO_SUM1_V
WHERE YEARCODE BETWEEN '200305' AND '200505'
GROUP BY SCCODE
SELECT  ROUND(AVG(AIV.BASICCNT)) AS BASICCNT
FROM
    (SELECT SUM(BASICCNT1)      AS BASICCNT
    FROM   ACCESS_INFO_SUM1_V
    WHERE YEARCODE BETWEEN '200305' AND '200505'
    GROUP BY sccode
     ) AIV
12 「"」 ①SELECT LENGTH('') AS VALUE1 FROM DUAL
[Result]VALUE1=NULL
①SELECT LENGTH('') AS VALUE1 FROM DUAL
[Result]VALUE1=0
②SELECT TO_DATE('','YYYYMMDD') AS VALUE2
FROM DUAL
[Result]VALUE2=NULL
②SELECT TO_DATE('','YYYYMMDD') AS VALUE2
FROM DUAL
[Result]VALUE2=0001-01-01 BC
③SELECT TO_NUMBER('',1) AS VALUE3 FROM DUAL
[Result]VALUE3=NULL
③SELECT TO_NUMBER('',1) AS VALUE3 FROM DUAL
[Result]不能
④INSERT INTO TEST(VALUE4)VALUES('')
[Result]VALUE4=NULL (注:VALUE3字段
型)
④INSERT INTO TEST(VALUE4)VALUES('')
[Result]VALUE4=0
  (注:VALUE4字段
型)
⑤INSERT INTO TEST(VALUE5)VALUES('')
[Result]VALUE5=NULL (注:VALUE5字段
字符 型)
⑤INSERT INTO TEST(VALUE5)VALUES('')
[Result]VALUE5=''
  (注:VALUE5字段
字符 型, 结果为长度为零的字符串 )
⑥INSERT INTO TEST(VALUE6)VALUES(TO_DATE('','YYYYMMDD'))
[Result]VALUE6=NULL (注:VALUE6字段
为时间类 型)
⑥INSERT INTO TEST(VALUE6)VALUES(TO_DATE('','YYYYMMDD'))
[Result]VALUE6=0001-01-01 BC
  (注:VALUE7字段
为时间类 型)
13 CEIL SELECT CEIL(SYSDATE - TO_DATE('20051027 14:56:10','YYYYMMDD HH24:MI:SS')) AS DAYS
FROM DUAL
SELECT
EXTRACT(DAY FROM (TO_TIMESTAMP(CURRENT_TIMESTAMP,'YYYY-MM-DD-HH24-MI-SS') -TO_TIMESTAMP('2005-10-27 14:56:10','YYYY-MM-DD-HH24-MI-SS') ))+1 AS DAYS
FROM DUAL
14 NULLIF 无 NULLIF函数 SELECT NULLIF(VALUE1,VALUE2) AS COL1 FROM DUAL
[注]当 VALUE1=VALUE2
时,COL1=NULL
15 CONCAT CONCAT(CHAR,CHAR) 创建函数来解决
CREATE FUNCTION CONCAT(CHAR,CHAR)
RETURNS CHAR AS
        'SELECT $1 || $2' LANGUAGE 'sql';
16 ADD_
MONTHS
add_months(date, int) 建函数来解决
CREATE FUNCTION add_months(date, int)
RETURNS date AS
'SELECT ($1 + ( $2::text || ''months'')::interval)::date;'
LANGUAGE 'sql'
17 LAST
_DAY
LAST_DAY(DATE) 建函数来解决
CREATE FUNCTION LAST_DAY(DATE)
RETURNS DATE AS
        'SELECT date(substr(text($1 +
                interval(''1 month'')),1,7)||''-01'')-1'
        LANGUAGE 'sql';
18 MONTHS
_BETWEEN
MONTH_BETWEEN(DATA,DATA) 建函数来解决
CREATE FUNCTION MONTH_BETWEEN(DATA,DATA)
RETURNS NUMERIC AS
        'SELECT to_number((date($1)-
                           date($2)),''999999999'')/31'
        LANGUAGE 'sql';
19 GRE~
ATEST
GREATEST (LEAST) 建函数来解决
CREATE OR REPLACE FUNCTION
   GREATEST(TEXT[]) RETURNS TEXT AS '
DECLARE
   ARRY ALIAS FOR $1;
   GREATEST TEXT;
BEGIN
   GREATEST := ARRY[1];
   FOR I IN 1 .. ARRAY_UPPER(ARRY,1) LOOP
      IF ARRY[I] > GREATEST THEN
        GREATEST := ARRY[I];
      END IF;
   END LOOP;
   RETURN GREATEST;
END;
' LANGUAGE 'PLPGSQL';

SELECT GREATEST( ARRAY['HARRY','HARRIOT','HAROLD'])
AS "Greatest";
20 BITAND BITAND(int,int) SELECT 値 & 値;
21 子条件   在FROM子条件中字段 须有列名,
处理方法用AS +别名
22 MINUS MINUS 以EXCEPT来替代
23 BIN_
TO_
NUM
SELECT BIN_TO_NUM(1,0,1,0) AS VALUE1 FROM DUAL SELECT CAST(B'1010' AS INTEGER) AS VALUE1
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