Question

几乎可以在SQL中创建如下所示的星形三角形。我知道这可以在任何其他编程语言（如C，C ++，Java）中轻松完成，但想知道它是否真的可以用于SQL或PL / SQL。我尝试使用Oracle中的双表来处理它，但无法通过它。

  *              *
 * *             * * 
* * *    or      * * *

如果有人知道的话，有人可以请一些。

Answer 1

最简单的方法就是这样。如果你想建立等边三角形而不是直角三角形，你可以变得更复杂。

SQL> ed
Wrote file afiedt.buf

  1  select rpad( '* ', level*2, '* ' )
  2    from dual
  3* connect by level <= 3
SQL> /

RPAD('*',LEVEL*2,'*')
--------------------------------------------------------------------------------
*
* *
* * *

Answer 2

不确定您要找的是什么。也许这个？

select '*' from dual
union all select '**' from dual
union all select '***' from dual

Example

Answer 3

试试这个..

 declare @x int,@y int,@diff int
 select @x=0,@y=10,@diff=2--diferrence between consecutive rows
 while @x<@y
 begin
    if @x=0 and @diff<>1
       print space((@y-@x)*@diff-1)+replicate('*',1)
    else if(@diff%2=0)
       print space((@y-@x)*@diff)+replicate('* ',@x+(@x*(@diff-1)))
    else
       print space((@y-@x)*@diff)+replicate('* ',@x+(@x*(@diff-1)))
    select @x=@x+1
 end

Answer 4

如果您想要的只是简单的三角形，那么您可以这样做：

SELECT '*' FROM table
UNION
SELECT '**' FROM table
UNION
SELECT '***' FROM table

Answer 5

declare @count int,@num int,@num1 int, @space int, @str varchar(50)
set @count = 5 set @num = 1
while(@num<=@count)
begin
  set @num1 = 0 set @space = @count-@num
  while (@num1<@num)
  begin
   if @str is null
    set @str = '* '
   else
    set @str = @str+'* '   set @num1 = @num1+1
  end
print (space(@space)+@str)
set @num = @num+1   set @str = null
end

Answer 6

[等式训练器]我们可以使用Oracle SQL进行金字塔构造，如下所示。

myTheme <- chart_theme()
myTheme$col$`bg` <- "#000000"

**这里的5是行数。

让我们扭转它，

select rpad(' ',5 -level) || rpad( '* ', level*2, '* ' )
   from dual
 connect by level <= 5;

Answer 7

这是在sql中获取完美三角形或金字塔的脚本（在Microsoft Sql 2008中测试）

declare @x int,@y int
select @x=5,@y=0
while @x>0
begin
print space(@x)+replicate('*',@y)+replicate('*',@y+1)
set @y=@y+1
set @x=@x-1
end

     *
    ***
   *****
  *******
 *********

你可以在这个链接上获得更多的脚本和帮助......这对我有帮助

链接： - sqlquerynscript

Answer 8

declare @row int = 5,
@index int = 0,
@string nvarchar(5) =''
while @row > 0
begin
    set @index = @row
    while @index > 0
    begin
        set @string = '*' + @string
        set @index = @index - 1
    end
    print @string
    set @string = ''
    set @row = @row - 1
end

 *****
 ****
 ***
 **
 *

Answer 9

DECLARE @lclMaxLevel INT=5
DECLARE @lclPrintCount INT =0

WHILE @lclMaxLevel > 0
  BEGIN
      PRINT Space(@lclMaxLevel)
            + Replicate('*', @lclPrintCount+1)

      SET @lclMaxLevel=@lclMaxLevel - 1
      SET @lclPrintCount=@lclPrintCount + 1
  END

Answer 10

select rpad('* ', level * 2, '* ')
  from dual connect by
    level <= 10
* 
* * 
* * * 
* * * * 
* * * * * 
* * * * * * 
* * * * * * * 
* * * * * * * * 
* * * * * * * * * 
* * * * * * * * * * 


select rpad(' ',r*2,' ')||rpad('* ',l*2,'* ') k 
  from ( select level l,row_number() over(order by null) r 
            from dual 
           connect by level<=10 
         order by l desc)

  * * * * * * * * * * 
    * * * * * * * * * 
      * * * * * * * * 
        * * * * * * * 
          * * * * * * 
            * * * * * 
              * * * * 
                * * * 
                  * * 
                    *



select rpad(' ',l*2,' ')||rpad('* ',r*2,'* ') k 
 from ( select level l,row_number() over(order by null) r 
          from dual 
         connect by level<=10 
         order by l desc)

                    * 
                  * * 
                * * * 
              * * * * 
            * * * * * 
          * * * * * * 
        * * * * * * * 
      * * * * * * * * 
    * * * * * * * * * 
  * * * * * * * * * * 

select rpad(' ',l,' ')||rpad('* ',r*2,'* ') k 
from ( select level l,row_number() over(order by null) r 
        from dual 
       connect by level<=10 
       order by l desc) 

      * 
     * * 
    * * * 
   * * * * 
  * * * * * 
 * * * * * * 
* * * * * * *

如何使用SQL打印星形三角形

10 个答案: