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Oracle和SQL Server分析挖掘函数

栏目:Oracle时间:2014-02-20 14:36:39
文中提及函数并非Oracle及SQL Server 的全部功能,尤其分析挖掘函数,并未完全涵盖,请以实际解决问题优先,勿妄谈二者优劣。

1.绝对值
  S:select abs(-1) value
  O:select abs(-1) value from dual

  2.取整(大)
  S:select ceiling(-1.001) value
  O:select ceil(-1.001) value from dual

  3.取整(小)
  S:select floor(-1.001) value
  O:select floor(-1.001) value from dual

  4.取整(截取)
  S:select cast(-1.002 as int) value
  O:select trunc(-1.002) value from dual

  5.四舍五入
  S:select round(1.23456,4) value 1.23460
  O:select round(1.23456,4) value from dual 1.2346

  6.e为底的幂
  S:select Exp(1) value 2.7182818284590451
  O:select Exp(1) value from dual 2.71828182

  7.取e为底的对数
  S:select log(2.7182818284590451) value 1
  O:select ln(2.7182818284590451) value from dual; 1

  8.取10为底对数
  S:select log10(10) value 1
  O:select log(10,10) value from dual; 1

  9.取平方
  S:select SQUARE(4) value 16
  O:select power(4,2) value from dual 16

  10.取平方根
  S:select SQRT(4) value 2
  O:select SQRT(4) value from dual 2

  11.求任意数为底的幂
  S:select power(3,4) value 81
  O:select power(3,4) value from dual 81

  12.取随机数
  S:select rand() value
  O:select sys.dbms_random.value(0,1) value from dual;

  13.取符号
  S:select sign(-8) value -1
  O:select sign(-8) value from dual -1

----------数学函数

  14.圆周率
  S:SELECT PI() value 3.1415926535897931
  O:不知道

  15.sin,cos,tan 参数都以弧度为单位
  例如:select sin(PI()/2) value 得到1(SQLServer)

  16.Asin,Acos,Atan,Atan2 返回弧度

  17.弧度角度互换(SQLServer,Oracle不知道)
  DEGREES:弧度-〉角度
  RADIANS:角度-〉弧度

  ---------数值间比较

  18. 求集合最大值
  S:select max(value) value from
  (select 1 value
  union
  select -2 value
  union
  select 4 value
  union
  select 3 value)a

  O:select greatest(1,-2,4,3) value from dual

  19. 求集合最小值
  S:select min(value) value from
  (select 1 value
  union
  select -2 value
  union
  select 4 value
  union
  select 3 value)a

  O:select least(1,-2,4,3) value from dual

  20.如何处理null值(F2中的null以10代替)
  S:select F1,IsNull(F2,10) value from Tbl
  O:select F1,nvl(F2,10) value from Tbl

  --------数值间比较

  21.求字符序号
  S:select ascii('a') value
  O:select ascii('a') value from dual

  22.从序号求字符
  S:select char(97) value
  O:select chr(97) value from dual

  23.连接
  S:select '11'+'22'+'33' value
  O:select CONCAT('11','22')||33 value from dual

23.子串位置 --返回3
  S:select CHARINDEX('s','sdsq',2) value
  O:select INSTR('sdsq','s',2) value from dual

  23.模糊子串的位置 --返回2,参数去掉中间%则返回7
  S:select patindex('%d%q%','sdsfasdqe') value
  O:oracle没发现,但是instr可以通过第四个参数控制出现次数
  select INSTR('sdsfasdqe','sd',1,2) value from dual 返回6

  
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