跟crackman做sas adv认证试题（3）-经管之家官网！

Given the SAS data set SASUSER.HIGHWAY:
Steering Seatbelt Speed Status Count
-------- -------- ----- ------- -----
absent No 0-29 serious 31
absent No 0-29 not 1419
absent No 30-49 serious 191
absent no 30-49 not 2004
absent no 50+ serious 216
The following SAS program is submitted:
proc sql noprint;
select distinct
Speed [_insert_SQL_clause_]
from SASUSER.HIGHWAY
;
quit;
title1 "Speed values represented are: &GROUPS";
proc print data=SASUSER.HIGHWAY;
run;
Which SQL clause stores the text 0-29,30-49,50+ in
the macro variable GROUPS?
A.into &GROUPS
B.into :GROUPS
C.into :GROUPS separated by ','
D.into &GROUPS separated by ','
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此题考察的是SQL与宏的结合。
用SQL语句产生宏变量，语法结构如下：
PROC SQL NOPRINT;
SELECT column1<,column2,...>
INTO :macro-variable-1<,:macro-variable-2,...>
FROM table-1 | view-1
<WHERE expression>
<other clauses>;
QUIT;
下面这个语法结构是将一变量的多个值组合在一个宏变量里面，例如：
General form, SELECT statement with INTO clause for combining values into one macro variable:
PROC SQL NOPRINT;
SELECT column1
INTO :macro-variable-1
SEPARATED BY ‘delimiter1’
FROM table-1 | view-1
<WHERE expression>
<other clauses>;
QUIT;
回头看看题目的要求：如何将这个文本“ 0-29,30-49,50+”放在一个宏变量&GROUPS中？
显然是采用第二个语法结构。
答案为C。
示例程序：
data crackman;
input Steering $ Seatbelt $ Speed $ Status $ Count@;
datalines;
absent No 0-29 serious 31
absent No 0-29 not 1419
absent No 30-49 serious 191
absent no 30-49 not 2004
absent no 50+ serious 216
;
run;
proc sql;
select distinct speed into:groups separated by ',' from crackman;
quit;
%put _all_;
大家可以思考一些，这里创建的宏变量是全局宏变量还是局部宏变量，这两者之间的差别是什么？