例如：62.98 242.17 74.51 253.72 5.11 184.27 82.29 261.47 71.69 250.90 59.70 238.89 58.83 238.01 28.70 207.85 106.46 285.65 62.56 241.71
第17个空格在该串的位置，即第17个空格到起始字符的字符个数。这个sql语句怎么写？
------解决方案--------------------
- SQL code
declare @str varchar(1000)declare @n intset @n=1select @str='62.98 242.17 74.51 253.72 5.11 184.27 82.29 261.47 71.69 250.90 59.70 238.89 58.83 238.01 28.70 207.85 106.46 285.65 62.56 241.71'while @n<=17beginselect @str=right(@str,len(@str)-charindex(' ',@str))select @[email protected]+1endprint @str
------解决方案--------------------
- SQL code
declare @sIncludeParaMDL varchar(1000),@nPosi int,@nPosi1 int,@nPosi2 intset @sIncludeParaMDL='62.98 242.17 74.51 253.72 5.11 184.27 82.29 261.47 71.69 250.90 59.70 238.89 58.83 238.01 28.70 207.85 106.46 285.65 62.56 241.71 'select @nPosi =charindex(' ',@sIncludeParaMDL,1)select @nPosi1=1select @nPosi2=1while @nPosi>0begin --这里是地几个空格 if(@nPosi2=18) select substring(@sIncludeParaMDL,1,@nPosi) --这里是第17个到18之间的数 if(@nPosi2=18) select substring(@sIncludeParaMDL,@nPosi1,@[email protected]) --select @nPosi,@nPosi1,@nPosi2 select @nPosi1 = @nPosi+1,@[email protected]+1 select @nPosi =charindex(' ',@sIncludeParaMDL,@nPosi1) end