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YadiAcho
Calcite | Level 5
Need help!!

I have data field look like this :

78687hkjkjh43@hsak-786778@hhjjjh762lkjklll

I want to get data beetween delimiter @.....@ I want get value hsak-786778
Thx for your help....

data getstr1(keep=memo test);
set Sastbl.sas;
test=input(delimiter="@" and "@",$10.);
run;

this can't solve my problem, any other solution please advice.thx....
4 REPLIES 4
Oleg
SAS Employee
Does not look nice, but formula should be
(SUBSTR(t1.A, Find(t1.a,'@')+1,Find(t1.a,'@',Find(t1.a,'@')+1)-Find(t1.a,'@')-1))

if You string contains in variable t1.A (I used EG for testing)
Hopefully You have not need to convert in this way million of rows.

P.S. I would add befoire this statement checks, that string really contains two pieces of @ sign.

/oleg
SPR
Quartz | Level 8 SPR
Quartz | Level 8
Hello YadiAcho,

This is another solution:
[pre]
data a;
i="78687hkjkjh43@hsak-786778@hhjjjh762lkjklll";
r=SCAN(i,2,'@');
run;
[/pre]
Sincerely,
SPR
YadiAcho
Calcite | Level 5
thx...

it works...
ChrisHemedinger
Community Manager
And as a general-purpose pattern-matching approach, regular expressions work great. Check out the PRXMATCH function.

http://www2.sas.com/proceedings/sugi29/129-29.pdf

Chris
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