Sas 删除最后一个逗号后的所有内容

我有一个地址字段，其地址类似于：

1 High Street, London, Uk, L1 6YR 
Flat 2, 35 Second Street, London,Greater London, L14 8FG

我想去掉最后一个逗号之后的所有内容，这样我就可以得到：

   1 High Street, London, Uk
   Flat 2, 35 Second Street, London,Greater London

我试着这样做：

DATA WANT;
SET HAVE;
ADDRESS_NEW = SCAN(ADDRESS, -1, ",");
RUN;

但这只是返回邮政编码：

L1 6YR
L14 8FG

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您可以使用

PRXCHANGE

修改符合所需条件的文本值。此示例使用regex模式

s/，[^，]*$/

将找到的结束部分替换为零

data have;
input address $80.;
datalines; 
1 High Street, London, Uk, L1 6YR 
Flat 2, 35 Second Street, London,Greater London, L14 8FG
Argy Bargy Lane
London, Uk
run;

data want;
  set have;
  address = prxchange("s/,[^,]*$//", 1, address);
run;

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具有B选项的FindC

24   data _null_;
25      input address $80.;
26      put / address=;
27      l = findc(address,',','b');
28      if l then address = substrn(address,1,l-1);
29      put  (_all_)(=);
30      datalines;


address=1 High Street, London, Uk, L1 6YR
address=1 High Street, London, Uk l=26

address=Flat 2, 35 Second Street, London,Greater London, L14 8FG
address=Flat 2, 35 Second Street, London,Greater London l=48

address=Argy Bargy Lane
address=Argy Bargy Lane l=0

address=London, Uk
address=London l=7

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还请分享您尝试过的任何代码，如果您遇到任何错误，请说明它们是什么。@Amir我已经添加了我认为应该尝试的内容。我不知道还有什么可以尝试的，所以我在这里发布了。