如何使用 SQL 打印星形三角形

How do I print a triangle of stars using SQL

提问by Teja

Is it practically possible to create a triangle of stars like this as below in SQL.I know that this could be done easily in any other programming language like C,C++,Java but want to know whether it is really possible with just SQL or PL/SQL.I tried working on it with dual table in Oracle but couldn't get through it.

实际上是否可以在 SQL 中创建如下所示的三角形三角形。我知道这可以在任何其他编程语言（如 C、C++、Java）中轻松完成，但想知道是否真的可以仅使用 SQL 或 PL /SQL。我尝试在 Oracle 中使用双表处理它，但无法通过它。

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

Can someone please shed somelight if anyone knows about it.

如果有人知道的话，有人可以透露一些信息。

回答by Justin Cave

The simplest approach would be something like this. You can get more sophisticated particularly if you want to build the equilateral triangle rather than the right triangle.

最简单的方法是这样的。如果您想构建等边三角形而不是直角三角形，则可以变得更加复杂。

SQL> ed
Wrote file afiedt.buf

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

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

回答by RedFilter

Not sure what you are looking for exactly. Perhaps this?

不知道你在找什么。也许这个？

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

Example

例子

回答by raja

Here is the script to Get a perfect triangle or Pyramid in sql (tested in Microsoft Sql 2008)

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

---

---

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

you can get many more scripts and help at this link... it was helpful to me

你可以在这个链接上获得更多的脚本和帮助......这对我很有帮助

Link:- sqlquerynscript

链接：- sqlquerynscript

回答by Benson

Try this..

尝试这个..

 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

回答by Anoop

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

回答by Justin Pihony

If all you want is the simple triangle, then you can do this:

如果你想要的只是简单的三角形，那么你可以这样做：

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

回答by Sameer Pradhan

[Equilateral Traingle] We can make a pyramid with Oracle SQL as follows.

【等边三角形】我们可以用Oracle SQL做一个金字塔如下。

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

** Here 5 is the number of lines.

** 这里 5 是行数。

Let us reverse it,

让我们反过来，

select rpad(' ',level) || rpad( '* ', 2*(5-level+1), '* ' )
  from dual
 connect by level <= 5; 

回答by helpMeLearn

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

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

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

回答by spandan

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

回答by Rajesh Dutta

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) 

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