About jhh197

jhh197 · ‎09-22-2018

I am so confused why below code doesn't work and y doesn't resolve to 2|3|1 %let xp1 = 2; %let xp2 = 3; %let xp3 = 1; %do i=1 %to 3; %if &i=1 %then %do; %let y = &&xp&i._crs; %end; %else %do; %let y =%sysfunc(catx(|,&y,&&xp&i._crs)); %end; %end; %put &y;

jhh197 · ‎09-22-2018

Hi Cynthia , Yes I have created it Thank you

jhh197 · ‎09-22-2018

hi I have tried this but I am getting like below it started with a pipe |2|3|1 thought it should be like this 2|3|1 Thank you

jhh197 · ‎09-22-2018

xp1_crs , xp2_crs and xp3_crs are already getting resolved in the macro and now I am using these values to prepare a list like 2|3|1 so the values 2 is the value of xp1_crs and value 3 is the value of macro variable xp2_crs and the value of 1 is the value of xp3_crs i am trying to concatenate macro variable values Thank you

jhh197 · ‎09-22-2018

Yes . the first code would work and can be used when we have three variables . But I wanted to do this in a do loop since I don't want to mention &x1 , &x2 , &x3 as I have 250 variables from x1 to x250 Thank you

jhh197 · ‎09-22-2018

oh the % let statements should be like below %let xp1_crs = 2; %let xp2_crs = 3; %let xp3_crs = 1;

jhh197 · ‎09-22-2018

I have couple of macro variables which has values and I want to create a list of those values for each variable into one and the final list should be %let xp1 = 2; %let xp2 = 3; %let xp3 = 1; %do i=1 %to 3; %if &i=1 %then %do; %let y = &&xp&i._crs; %end; %else %do; %let y =%sysfunc(catx(|,&y,&&xp&i._crs)); %end; %end; %put &y; so value of y should be 2|3|1 after the loop is completed Can anyone please help

jhh197 · ‎04-25-2018

Hi , I have a string %let Psv = DISC_CP|DISC_SP|DISC_MI|AMECS_CP|AMECS_SP|AMECS_MI; So I have a dataset test1 and wanted to create test2 and test3 Test1 MEMID ITEMID DISC_CP DISC_SP DISC_MI AMECS_CP AMECS_SP AMECS_MI 501489 25689 2.3 1.2 3.3 6.7 2.2 5.6 501489 23287 2.3 1.2 3.3 6.7 2.4 5.6 501489 26625 2.6 1.3 3.3 6.8 2.4 5.3 702569 32589 3.5 1.4 4.2 8.9 1.2 3.1 702569 32291 3.5 1.4 4.2 8.7 1.2 3.1 702569 38896 2.1 1.5 4.2 8.3 1.2 4.9 So if the string &Psv contains DISC then it should create Test2 dataset which looks like below group by MEMID AND ITEMID MEMID ITEMID DISC_CP DISC_SP DISC_MI 501489 25689 2.3 1.2 3.3 501489 26625 2.6 1.3 3.3 So if the string &Psv contains AMECS then it should create Test3 dataset which looks like below group by MEMID only MEMID AMECS_CP AMECS_SP AMECS_MI 501489 6.73333 2.3333 5.5 702569 8.63333 1.2 3.7 Can anyone please help Thanks

jhh197 · ‎01-25-2018

I am running SAS EG 71 64 bit Thank you

jhh197 · ‎01-24-2018

I have two excel files test1 and test2 test1.xlsx has three tabs as shown below test1.xlsx has 3 tabs APPLE A B C 123 4 5 2 2.3 4.5 orange A B C D 123 4 5 6 2 2.3 4.5 7 grape A B C 123 4 5 2 2.3 4.5 TEST2.XLSX has two tabs apple A B C 55 55 66 2 2.3 4.5 orange A B C D 46 36 59 6 2 2.3 4.5 7 and after appending Test1 and Test2.xlsx my final excel file test3.xlsx should look like below with 3 tabs APPLE A B C 123 4 5 2 2.3 4.5 55 55 66 2 2.3 4.5 ORANGE A B C D 123 4 5 6 2 2.3 4.5 7 46 36 59 6 2 2.3 4.5 7 grape A B C 123 4 5 2 2.3 4.5 Can anyone please help how do I append two excel files ?

jhh197 · ‎10-20-2017

This also helped Thank you so much for quick response Thank you 🙂

jhh197 · ‎10-20-2017

Hi , Thank you this worked for detecting comma position and counting the arguments Thank you 🙂

jhh197 · ‎10-20-2017

Sorry I used this macro for count %let h= %sysfunc(countw(&u,',')); %put &h;

jhh197 · ‎10-20-2017

Yes I used %let x = index(&u, ','); % put &x but it's not resolved Error :the quoted string currently being processed became more than 262 characters long . The string might have unbalanced quotation marks quotation For count I tried % let h = %sysfunc(countw(&x,',')); %put &h;

jhh197 · ‎10-20-2017

Hi I have macro variable %let u = 6,22,5 I want a macro code to check whether comma exists in the macro u if yes then I want a syntax to count macro u ( Answer should be 3 in above case ) next I want syntax to extract 6 into one macro 22 into one macro 5 into one macro variable Thank you

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