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标题: 跟crackman做sas adv认证试题（6） [打印本页]
作者: shiyiming    时间: 2010-11-16 16:02
标题: 跟crackman做sas adv认证试题（6）
From crackman's blog on Baidu


               
                <p><font size="4" face="宋体"> Given the SAS data set WORK.TRANSACT:  <br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  Rep&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  Cost&nbsp;&nbsp;  Ship&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  -----&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  -----  -----&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  200&nbsp;&nbsp;&nbsp;&nbsp;  50&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  400&nbsp;&nbsp;&nbsp;&nbsp;  20&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  JONES&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;&nbsp;&nbsp;  10&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  600&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  JONES&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  5&nbsp;&nbsp;  <br>
<br>
  The following output is desired:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  Rep&nbsp;&nbsp;&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  -----&nbsp;&nbsp;  ----&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  JONES&nbsp;&nbsp;&nbsp;  105&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;  250&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
<br>
  Which SQL statement was used? </font></p>
<p><br>
<font size="4" face="宋体">A.select rep,min(Cost+Ship)<br>
from WORK.TRANSACT order by Rep  <br>
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
<br>
B.select Rep,min(Cost,Ship) as Min<br>
from WORK.TRANSACT&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
summary by Rep order by Rep  <br>
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
<br>
C.select Rep,min(Cost,Ship)<br>
from WORK.TRANSACT&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
group by Rep order by Rep  <br>
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
<br>
D.select  Rep,min(Cost+Ship)<br>
from WORK.TRANSACT&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
group by Rep order by Rep  <br>
;<br>
答案是 D。<br>
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答案A:<br>
得出的结果应该是没有分组比较的结果，就是自身比较，所以会得到5个结果，没有分组，而且MIN的值都为105<br>
答案B:<br>
语法错误，得不到结果<br>
答案C:<br>
得到是在一个REP上的这两个变量COST SHIP之间的小值。<br>
答案D：正确答案<br>
因为在要得到的数据集中有一个是没有命名的，所以在<br>
select  Rep,min(Cost+Ship)<br>
from WORK.TRANSACT&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
group by Rep order by Rep  <br>
;<br>
中MIN(COST+SHIP)是没有命名的变量的值。<br>
示例程序：<br>
data TRANSACT;<br>
input Rep $  Cost Ship@;<br>
datalines;<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  200&nbsp;&nbsp;&nbsp;&nbsp;  50&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  400&nbsp;&nbsp;&nbsp;&nbsp;  20&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  JONES&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;&nbsp;&nbsp;  10&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  SMITH&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  600&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;  <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  JONES&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  100&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  5  <br>
&nbsp;&nbsp;  ;<br>
&nbsp;&nbsp;  run;<br>
proc sql;<br>
select rep,min(Cost+Ship)<br>
from WORK.TRANSACT order by Rep ;<br>
select Rep,min(Cost,Ship)<br>
from WORK.TRANSACT&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  <br>
group by Rep order by Rep  <br>
;  <br>
quit;</font></p> <a href="http://hi.baidu.com/crack%5Fman/blog/item/68a5fadaf971f34496ee3763.html">阅读全文</a>
               
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