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sqlserversubstring函数使用方法小结 - 站长搜索
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sqlserversubstring函数使用方法小结

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2022-09-23 16:54:03 转载来源: 网络整理/侵权必删

在操作sqlserver时候用到了substring函数SUBSTRING(expression,start,length)参数expression字符串、二进制字符串、文本、图像、列或包含列的表达式。请勿使用包含聚合函数的表达式

在操作sqlserver时候用到了substring函数
SUBSTRING ( expression, start, length )

参数
expression
字符串、二进制字符串、文本、图像、列或包含列的表达式。请勿使用包含聚合函数的表达式。

start
整数或可以隐式转换为 int 的表达式,指定子字符串的开始位置。

length
整数或可以隐式转换为 int 的表达式,指定子字符串的长度(但是结果并补一个返回length字符长度,要看expression的长度和start的设置!!!!!!!!!!!)。

返回值
如果 expression 是一种支持的字符数据类型,则返回字符数据。如果 expression 是一种支持的二进制数据类型,则返回二进制数据。如果 start = 1,则子字符串从表达式的第一个字符开始

需要补充一句的是,开始start位置的设置,确实如果设置为1,会从第一个字符开始计数,2,3,4....以此类推、

select substring(ziduan,1,6)

将会选择字段的前6个字符,比如说是200908

如果设置为0,

同样的执行select substring(ziduan,0,6)

结果会是20090,如果换成-1

同样执行select substring(ziduan,-1,6)

结果会是2009

同样-2,-3,.................依次类推
下面是网络上的其它一些文章。
substring('98765',-1,3) 结果:9
substring('98765',0,3) 结果:98
substring('98765',1,3) 结果:987
关于函数的这种情况,找了点资料,但是仍不是很明白
SubString($sourceString ,$startingLoc,$length)
$sourceString
资源字符串。

$startingLoc
子字符串在资源字符串中的起点。如果此值为负数或 0,则只返回那些所在位置大于零的字符。如果此值大于 $sourceString 的长度,则返回长度为零的字符串。

$length
[可选] 要检索的字符数。如果没有指定此值,它将返回从 $startingLoc 中指定的位置开始直到字符串末尾的所有字符。

如果此值为负数或 0,则只返回那些所在位置大于零的字符。

入门学习SQL
select substring(convert(varchar,birthdate),5,1) as mon ,* from employees
where substring(convert(varchar,birthdate),5,1)='8'

标签: sqlserversubstring 函数 使用方法 小结


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