Will query A and query B yield same result below?   Will these two queries yield same results even if the data changes? Is the left outer join along with where condition in Query A is equivalent to the inner join in Query B for all cases of data. Are there any cases for which the two queries will not be identical? Can Query A be replaced with Query B to get same results?   Please advise.   Query A proc sql; select a.key, a.val1, b.val2, c.val3 from a left join b On a.key=b.key left join c on a.key=c.key where b.val2=20 and c.val3=30 ; quit;   Query B proc sql; select a.key, a.val1, b.val2, c.val3 from a inner join b on a.key=b.key and b.val2=20 inner join c on a.key=c.key and c.val3=30 ; quit;   Data: Sample data below, you may change cases to break the query if you need to. data a; length key $1. val1 8.; input key val1; datalines; a 0 a 10 b 10 b 20 c 30 d 40 e 50 ; run; data b; length key $1. val2 8.; input key val2; datalines; b 20 b 10 c 10 d 20 e 20 f 20 g 10 g 20 ; run; data c; length key $1. val3 8.; input key val3; datalines; a 10 b 20 b 30 c 20 c 30 c 40 d 30 e 50 h 20 h 30 ; run; ... View more

Hi everyone, This is my first post with SAS Support communities, so please bear with me and pardon my misgivings. I ran into an issue lately where I was trying to convert a character variable into a numeric variable while extracting data from a dataset. The character variable was a 18 digit long numeric value stored as character. So, when I tried importing the variable and converting it into numeric using the input function, only the first 16 digits were getting converted successfully and the last 2 digits were getting arbitrarily random values. So, I took the empirical approach and devised the following simple code to check is SAS is able to process the numericals upto 10^18 numerical precision, and I found that last 2 digits were always lost. Can you please help me understand if my conclusion is correct? Is SAS unable to handle numerical variables beyond 10^16 value? Please adivse. Code: 214  data _null_ ; 215     do i=1 to 17 ; 216        big_num = 10**i + 1 ; 217        put big_num best18. +2 i 2. ; 218     end ; 219  run ; Result:                 11   1                101   2               1001   3              10001   4             100001   5            1000001   6           10000001   7          100000001   8         1000000001   9        10000000001  10       100000000001  11      1000000000001  12     10000000000001  13    100000000000001  14   1000000000000001  15 10000000000000000  16 100000000000000000  17 For observation 16 and 17 it can be seen that the numerical precision is lost. Please advise how can I go around this issue? Thanks in advance. ... View more