PostgreSQL与Oracle的区别

2024年4月20日 142次阅读 来源: PostgreSQL
NO问题点OraclePostgreSQL
1DUALSELECT 1+1 FROM DUALSELECT 1+1
或者
CREATE VIEW dual AS
      SELECT ‘X’::VARCHAR(1) AS DUMMY
再 SELECT 1+1 FROM DUAL
2NEXTVALSELECT A_TABLE_SEQUENCE.NEXTVAL
FROM   DUAL		SELECT NEXTVAL(‘A_TABLE_SEQUENCE’)
FROM   DUAL
3ROWNUM①SELECT *
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)
5ASSELECT 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
6NVLSELECT 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 
7TO_
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位数为COL1字段的长度]
8DECODESELECT 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
11aggregateSELECT 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]不能执行
④INSERT INTO TEST(VALUE4)VALUES(”)
[Result]VALUE4=NULL (注：VALUE3字段为数值类型)		④INSERT INTO TEST(VALUE4)VALUES(”)
[Result]VALUE4=0
 (注：VALUE4字段为数值类型)
⑤INSERT INTO TEST(VALUE5)VALUES(”)
[Result]VALUE5=NULL (注：VALUE5字段为字符类型)		⑤INSERT INTO TEST(VALUE5)VALUES(”)
[Result]VALUE5=”
 (注：VALUE5字段为字符类型,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)
13CEILSELECT 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
14NULLIF无NULLIF函数SELECT NULLIF(VALUE1,VALUE2) AS COL1 FROM DUAL
[注]当VALUE1=VALUE2时，COL1=NULL
15CONCATCONCAT(CHAR,CHAR)创建函数来解决
CREATE FUNCTION CONCAT(CHAR,CHAR)
RETURNS CHAR AS
        ‘SELECT $1 || $2’ LANGUAGE ‘sql’;
16ADD_
MONTHS		add_months(date, int)创建函数来解决
CREATE FUNCTION add_months(date, int)
RETURNS date AS
‘SELECT ($1 + ( $2::text || ”months”)::interval)::date;’
LANGUAGE ‘sql’
17LAST
_DAY		LAST_DAY(DATE)创建函数来解决
CREATE FUNCTION LAST_DAY(DATE)
RETURNS DATE AS
        ‘SELECT date(substr(text($1 +
                interval(”1 month”)),1,7)||”-01”)-1′
        LANGUAGE ‘sql’;
18MONTHS
_BETWEEN		MONTH_BETWEEN(DATA,DATA)创建函数来解决
CREATE FUNCTION MONTH_BETWEEN(DATA,DATA)
RETURNS NUMERIC AS
        ‘SELECT to_number((date($1)-
                           date($2)),”999999999”)/31′
        LANGUAGE ‘sql’;
19GRE~
ATEST		GREATEST (LEAST)创建函数来解决
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”;

20BITANDBITAND(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
22MINUSMINUS以EXCEPT来替代
23BIN_
TO_
NUM		SELECT BIN_TO_NUM(1,0,1,0) AS VALUE1 FROM DUALSELECT CAST(B’1010′ AS INTEGER) AS VALUE1
    原文作者：PostgreSQL
    原文地址: https://www.cnblogs.com/sekihin/archive/2009/01/14/1375329.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
点赞