如何对宏变量进行替换处理

%macro aaa(y=);
%let x=\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42;

%put &x;

%mend
y指定为1, 将宏变量的第1个\aaa\bbb替换为空，x解析为\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42；
y指定为2, 将宏变量的前2个\aaa\bbb替换为空，x解析为\ccc21\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42；

1. %macro aaa(y=);
   
2. %let x=\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42;
   
3. %let expressid=%sysfunc(prxparse(s/\\aaa\\bbb//));
   
4. %let x=%sysfunc(prxchange(&expressid,&y,%str(&x)));
   
5. %put &x;
   
6. %mend;
   
7. 
   
8. %aaa(y=1)
   
9. %aaa(y=2)

复制代码

学习学习

%macro aaa(y=);
%let x=\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42;

%do i=1 %to &y;
     %let pos=%index(&x,\aaa\bbb);
     %if &pos=1 %then %let x=%substr(&x,9);
         %else %if &pos %then %let x=%substr(&x,1,&pos-1)%substr(&x,&pos+8);
%end;

%put &x;
%mend;

%macro want(y=);
%let x=%str(\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42);
data a;
        c=count("&x","\aaa\bbb");
        do i=&y to &y;
                d=compress(tranwrd(substr("&x",1,find("&x","\aaa\bbb",i)-1),"\aaa\bbb",""))||
                        substr("&x",find("&x","\aaa\bbb",i)+8);
        end;
run;
%mend;
%want(y=2);

你可以根据你的需要，对宏程序中的函数进行调整

huangpengfei 发表于 2014-11-2 22:09
%macro want(y=);
%let x=%str(\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42);
data a;

提供了另外一种思路，但程序是有bug的

pobel 发表于 2014-11-2 10:57
%macro aaa(y=);
%let x=\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42;

\aaa\bbb之间字符的长度不固定

huangpengfei 发表于 2014-11-2 22:09
%macro want(y=);
%let x=%str(\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42);
data a;

\aaa\bbb\ 之间字符的长度不固定

谢谢！

pobel 发表于 2014-11-2 10:57
%macro aaa(y=);
%let x=\aaa\bbb\ccc21\aaa\bbb\ccc23\aaa\bbb\ccc35\aaa\bbb\ccc42;

%do i=1 %to &y;
     %let pos=%index(&x,\aaa\bbb);
     %if &pos=1 %then %let x=%substr(&x,9);
         %else %if &pos %then %let x=%substr(&x,1,&pos-1)%substr(&x,&pos+8);
%end;

%put &x;
%mend;

谢谢！