Oracle中的INSTR,NVL和SUBSTR函数的用法详解

Oracle中INSTR的用法:

INSTR方法的格式为

INSTR(源字符串, 要查找的字符串, 从第几个字符开始, 要找到第几个匹配的序号)

返回找到的位置,如果找不到则返回0.

例如:INSTR('CORPORATE FLOOR','OR', 3, 2)中,源字符串为'CORPORATE FLOOR', 在字符串中查找’OR’,从第三个字符位置开始查找”OR”,取第三个字后第2个匹配项的位置。

默认查找顺序为从左到右。当起始位置为负数的时候,从右边开始查找。

所以SELECT INSTR('CORPORATE FLOOR', 'OR', -1, 1) "aaa" FROM DUAL的显示结果是

Instring
——————
14

 oracle的substr函数的用法:

 取得字符串中指定起始位置和长度的字符串   substr( string, start_position, [ length ] )

 如: 

substr(‘This is a test’, 6, 2) would return ‘is’
substr(‘This is a test’, 6) would return ‘is a test’
substr(‘TechOnTheNet’, -3, 3) would return ‘Net’
substr(‘TechOnTheNet’, -6, 3) would return ‘The’
select substr(‘Thisisatest’, -4, 2) value from dual

 综合应用:

SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, -1, 1) “Instring” FROM DUAL
–INSTR(源字符串, 目标字符串, 起始位置, 匹配序号)
SELECT INSTR(‘CORPORATE FLOOR’,’OR’, 3, 2) “Instring” FROM DUAL
SELECT INSTR(‘32.8,63.5′,’,’, 1, 1) “Instring” FROM DUAL
SELECT SUBSTR(‘32.8,63.5’,INSTR(‘32.8,63.5′,’,’, 1, 1)+1) “INSTRING” FROM DUAL
SELECT SUBSTR(‘32.8,63.5’,1,INSTR(‘32.8,63.5′,’,’, 1, 1)-1) “INSTRING” FROM DUAL
— CREATED ON 2008-9-26 BY ADMINISTRATOR
DECLARE
— LOCAL VARIABLES HERE
T VARCHAR2(2000);
S VARCHAR2(2000);
NUM INTEGER;
I INTEGER;
POS INTEGER;
BEGIN
— TEST STATEMENTS HERE
T := ‘12.3,23.0;45.6,54.2;32.8,63.5;’;
SELECT LENGTH(T) – LENGTH(REPLACE(T, ‘;’, ”)) INTO NUM FROM DUAL;
DBMS_OUTPUT.PUT_LINE(‘NUM:’ || NUM);
POS := 0;
FOR I IN 1 .. NUM LOOP
DBMS_OUTPUT.PUT_LINE(‘I:’ || I);
DBMS_OUTPUT.PUT_LINE(‘POS:’ || POS);
DBMS_OUTPUT.PUT_LINE(‘==:’ || INSTR(T, ‘;’, 1, I));
DBMS_OUTPUT.PUT_LINE(‘INSTR:’ || SUBSTR(T, POS + 1, INSTR(T, ‘;’, 1, I) – 1));
POS := INSTR(T, ‘;’, 1, I);
END LOOP;
END;
— Created on 2008-9-26 by ADMINISTRATOR
declare
— Local variables here
i integer;
T VARCHAR2(2000);
S VARCHAR2(2000);
begin
— Test statements here
–历史状态
T := ‘12.3,23.0;45.6,54.2;32.8,63.5;’;
IF (T IS NOT NULL) AND (LENGTH(T) > 0) THEN
–T := T || ‘,’;
WHILE LENGTH(T) > 0 LOOP
–ISTATUSID := 0;
S := TRIM(SUBSTR(T, 1, INSTR(T, ‘;’) – 1));
IF LENGTH(S) > 0 THEN
DBMS_OUTPUT.PUT_LINE(‘LAT:’||SUBSTR(‘32.8,63.5’,1,INSTR(‘32.8,63.5′,’,’, 1, 1)-1));
DBMS_OUTPUT.PUT_LINE(‘LON:’||SUBSTR(‘32.8,63.5’,INSTR(‘32.8,63.5′,’,’, 1, 1)+1));
— COMMIT;
END IF;
T := SUBSTR(T, INSTR(T, ‘;’) + 1);
END LOOP;
END IF;
end;

PS:下面看下Oracle中INSTR、SUBSTR和NVL的用法

 INSTR用法:INSTR(源字符串, 要查找的字符串, 从第几个字符开始, 要找到第几个匹配的序号)

   返回找到的位置,如果找不到则返回0. 默认查找顺序为从左到右。当起始位置为负数的时候,从右边开始查找。若起始位置为0,返回值为0。

SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, 0, 1) FROM DUAL; 返回值为0
SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, 2, 1) FROM DUAL; 返回值为2
SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, 2, 2) FROM DUAL; 返回值为5
SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, -1, 1) FROM DUAL; 返回值为14
SELECT INSTR(‘CORPORATE FLOOR’, ‘OR’, -5, 1) FROM DUAL; 返回值为5

SUBSTR用法:SUBSTR( 源字符串, 查找起始位置, [ 长度 ] )

   返回值为源字符串中指定起始位置和长度的字符串。

SELECT SUBSTR(‘This is a test’, 0, 2) value from dual; 返回值Th
SELECT SUBSTR(‘This is a test’, 1, 2) value from dual; 返回值Hi
SELECT SUBSTR(‘This is a test’, -1, 2) value from dual; 返回值t
SELECT SUBSTR(‘This is a test’, -2, 2) value from dual; 返回值st

NVL用法:NVL(eExpression1, eExpression2)

   从两个表达式返回一个非 null 值。如果eExpression1的计算结果为null值,则 NVL( ) 返回eExpression2。如果eExpression1的计算结果不是null值,则返回eExpression1。eExpression1 和eExpression2可以是任意一种数据类型。如果eExpression1与eExpression2 的结果皆为 null值,则NVL( )返回NULL。

SELECT nvl(‘pos1′,null) from dual; 返回值为pos1
SELECT nvl(null,’pos2’) from dual; 返回值为pos1
SELECT nvl(null,null) from dual; 返回值为null

以上所述是小编给大家介绍的Oracle中的INSTR,NVL和SUBSTR函数的用法详解,大家如有疑问可以留言,或者联系站长。感谢亲们支持!!!


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