10-05-2015 04:51 AM
10-05-2015 05:31 AM
Please clarify what is not expected, provide some example strings in the form of a datastep and where they do not appear as you expect. The help on each of these functions is quite clear:
And I do not see why you would use the two functions together, unless in concatentaion. Why not just use strip()?
10-05-2015 06:49 AM
10-05-2015 06:58 AM
Well, cant really tell from that. I ran these two and the output was the same $ab_cd$. Were you using different operating systems, versions of SAS, was it the same data, same options etc. Could be any number of things. For the code though, I would recommend using strip() as its simpler, and avoids reading the code differently. However, an even more simpler form would be cats():
data temp; x="ab_cd "; y="$"||trim(left(x))||"$"; z="$"||left(trim(x))||"$"; v="$"||strip(x)||"$"; w=cats("$",x,"$"); run;
10-05-2015 09:00 AM
10-05-2015 09:25 AM
You should expect different results. The correct version is trim(left(value)). Consider what happens if you try left(trim(value)). First, the TRIM function removes trailing blanks. Then the LEFT function takes the leading blanks and puts them at the end of the string. Better yet, take a look at the STRIP function that removes both leading and trailing blanks.
10-05-2015 02:52 PM
Actually I got the reason.
if we store the results in a variable a and apply any appending special character causing the same output.
If we assign the values to a variable results to write in data set and filling back the length by 8 characters hence same output.
And if apply appending a special character before storing to any variable (writing back to data set) results the different out put.
For above explanation,
10-05-2015 04:18 PM
Hi. Be careful with character variable lengths.
The results of using character functions all have default lengths.
If you run PROC CONTENTS on data set TEMP, you'll see that the variable W created with the CATS has a length of 200 and that none of the other functions (TRIM or STRIP) change the length of the resulting variables V, Y, and Z.
The lengths will still be the length of any variables involved in the concatenation plus the lengths of any character constants. In this example that's 8, not because that's the default lenght of a character variabe, but because it's 6 (length of X) plus 2 (those two $);
10-06-2015 05:03 AM