About sagulolo

sagulolo · ‎03-16-2020

Dear all, Today i have tested the vbscript code below and it run without any issue, not sure why it failed at the begining. prjPaths = Array("C:\Users\XXXXXX\Desktop\NewSchedule\PROJECT 1.egp", "C:\Users\XXXXXX\Desktop\NewSchedule\PROJECT 2.egp") For Each prjName in prjPaths Hope this help others. Thanks for your time

sagulolo · ‎03-10-2020

Dear all, I'm have successfully schedule my program with EG, plus I getting idea from https://communities.sas.com/t5/SAS-Enterprise-Guide/Running-SAS-EG-Projects-sequentially/td-p/342843 to schedule few project in a sequence that I want by amended the vbscript, please see below which the schedule job run without any issue(at least in my view the task is perform without issue) From prjName = "C:\Users\XXXXXX\Desktop\Project1.egp" 'Project Name To prjPaths = Array("C:\Users\XXXXXX\Desktop\NewSchedule\PROJECT1.egp", "C:\Users\XXXXXX\Desktop\NewSchedule\PROJECT2.egp") For Each prjName in prjPaths The only issue for me is i have project which the name have space or path with space, like path = C:\Users\XXXXXX\Desktop\New Schedule project name = PROJECT 1.egp I have try to google but no luck and trying the code such as """C:\Users\XXXXXX\Desktop\NewSchedule\PROJECT 1.egp""" "C:\Users\XXXXXX\Desktop\NewSchedule\""PROJECT 1.egp""" Appreciated if you could guide/advise me the correct syntax to fix the above, thank you for your time and thank in advance.

sagulolo · ‎08-20-2019

Dear @hashman @KachiM , My deepest apologies for delay as I couldn't make it last Friday before I have my long leave. Below is my tested results, 1486 observations read from the data set WORK.TABLE_A. 1007 observations read from the data set WORK.TABLE_B. Total of 2493 observations only. Hash Method: NOTE: There were 1486 observations read from the data set WORK.TABLE_A. NOTE: There were 1007 observations read from the data set WORK.TABLE_B. NOTE: The data set WORK.WANT has 1085 observations and 5 variables. NOTE: DATA statement used (Total process time): real time 0.03 seconds cpu time 0.03 seconds Array Method: NOTE: There were 1486 observations read from the data set WORK.TABLE_A. NOTE: There were 1007 observations read from the data set WORK.TABLE_B. NOTE: The data set WORK.WANT has 1085 observations and 5 variables. NOTE: DATA statement used (Total process time): real time 0.01 seconds cpu time 0.01 seconds From the above time, I can say the array method is faster than Hash method. However, i wish i can choose both as solution because both also produce the same observation numbers(sorry didn't have time to check all), Hope this will benefit others also, I'm also welcome others expert to provide other solution and comments as well. Thank again for both your time.

sagulolo · ‎08-08-2019

@KachiM Thank for another tips - Data step debugger as this is first time I hear this, just have a googling on this and this is way better visual than to remove the noprint and cancel those drop to understand the flow. @hashman Deepest apologies for late reply, when I test the first solution(array), you have already share the 2nd solution(hash method) and even further explain the logical idea of solving this puzzle. The array was work perfectly, I will test the 2nd solution tomorrow and will compare the time consume for both solution. Then I will share the time spend for each solution and to confirm the best solution(actually not sure if i doing the fair comparison, thinking of using the same real data and just compare the time spend). Best Regards

sagulolo · ‎08-06-2019

@hashman Thank you so much for the coding, this is what I need and I never though of array can solve this puzzle. Now I have to test it at my real data, nevertheless apologize for my bad english. I will update here again after I test it with the real data. @KachiM Really appreciate your giving comments on the lack of logic explanations and guiding me to write a explanation to get the desire result. I also don't mind if you share others solution as I still new and lack of logical mind to resolve issue/puzzle like this. Best regards

sagulolo · ‎08-02-2019

Hi datasp, I'm okay with the data step approach and really thank for looking into this matter. Thank you so much

sagulolo · ‎08-02-2019

Hi datasp, 5th row in 500 - 5, how do you get this 5? See In Table_B, k12 is 20. I get the output 5 is because 20 - 10 - 5 (KEY12 - KEY9 - KEY10) = 5, so need another 5 from KEY 11 to become zero. Thank you

sagulolo · ‎08-02-2019

Hi datasp, The desire output will be as per below:- Sg_a Key_a Key_b actual_transfer A KEY1 KEY3 10 A KEY1 KEY4 25 A KEY1 KEY5 15 A KEY2 KEY5 20 B KEY6 KEY7 200 B KEY6 KEY8 700 C KEY9 KEY12 10 C KEY10 KEY12 5 C KEY11 KEY12 5 C KEY11 KEY13 400 C KEY11 KEY14 10 C KEY11 KEY15 50 Thank you

sagulolo · ‎08-02-2019

Hi datasp, Apologize for late reply. What is happening the next A of Table1? no action require as Table_B no more A group Is it to be ignored and go to B of Table 1? Yes, next will be cross match between Table_A and Table_B for B group. Added balance_extra into table for more understanding, thank you Sg_a Key_a Key_b Balance_Extra actual_transfer A KEY1 KEY3 50 -10 = 40 10 A KEY1 KEY4 40 - 25 = 15 25 A KEY1 KEY5 15 - 15 = 0(Key 1 no more number, next use Key 2) 15 A KEY2 KEY5 80 - 20, stop as table_B no more group for A 20 B KEY6 KEY7 1000 - 200 = 800 200 B KEY6 KEY8 800 - 700 = 100, stop as table_B no more group for B 700 C KEY9 KEY12 10 - 10 = 0, (Key 9 no more number, next use Key 10) 10 C KEY10 KEY12 5 - 5 = 0, (Key 10 no more number, next use Key 11) 5 C KEY11 KEY12 500 - 5 = 495 5 C KEY11 KEY13 495 - 400 = 95 400 C KEY11 KEY14 95 - 10 = 85 10 C KEY11 KEY15 85 - 50 = 35, stop for as table_B no more group for C 50

sagulolo · ‎08-01-2019

Hi datasp, Apologize as didn't really answer your question. "actual_transfer" is refer to sg_b if table A extra more than sg_b short. Else will be sg_a balance left. Once the table B short all deducted then move to next B then C.

sagulolo · ‎08-01-2019

Hi datasp, Thank for your time and sorry for my bad english, please refer below:- A KEY1(table A) --> extra 50 - A KEY3(table B) short 10 (50 - 10, can offset 10), so i get below:- so i output A KEY1(table A) A KEY3(table B) 10, then A KEY1(table A) -> still have extra $40 A KEY1(table A) --> extra 40 - A KEY4(table B) short 25 (40 - 25, can offset 25) so i output A KEY1(table A) A KEY4(table B) 25, then A KEY1(table A) -> still have extra 15 A KEY1(table A) --> extra 15 - A KEY5(table B) short 35 (15 - 35, only can offset 15) so i output A KEY1(table A) A KEY5(table B) 15, then next i refer to A KEY2 80 for the balance of A KEY5(table B) which is 20 A KEY2 (table A) -->extra 80 - A KEY5(table B) short 20 ( 80 - 20, can offset 20) so i output A KEY2 (table A) A KEY5(table B) short 20 hope the above clarifies.

sagulolo · ‎08-01-2019

Hi all, i have 2 table which is Table_A and Table_B, Table_A extra number will always sufficient to cover the short number at Table_B. The logic will be use the extra number from Table A to knock off the first short number from Table B with to the same group(Sg_a = Sg_b) until zero then follow by second short number from the same group. Please refer below:- data Table_A; infile datalines dlm='|'; input Sg_a $ Key_a $ extra; ; datalines; A |KEY1 |50 A |KEY2 |80 B |KEY6 |1000 C |KEY9 |10 C |KEY10 |5 C |KEY11 |500 ; data Table_B; infile datalines dlm='|'; input Sg_b $ Key_b $ short; ; datalines; A |KEY3 |10 A |KEY4 |25 A |KEY5 |35 B |KEY7 |200 B |KEY8 |700 C |KEY12 |20 C |KEY13 |400 C |KEY14 |10 C |KEY15 |50 ; proc sql; create table rightjoin as select a.*, b.* from Table_A a right join Table_B b on a.Sg_a = b.Sg_b; quit; So far, I have tried to right join table A to table B, but have no idea how to evaluate and keep track the actual_transfer, below is my desire output. Sg_a Key_a Key_b actual_transfer A KEY1 KEY3 10 A KEY1 KEY4 25 A KEY1 KEY5 15 A KEY2 KEY5 20 B KEY6 KEY7 200 B KEY6 KEY8 700 C KEY9 KEY12 10 C KEY10 KEY12 5 C KEY11 KEY12 5 C KEY11 KEY13 400 C KEY11 KEY14 10 C KEY11 KEY15 50 Appreciate your time and help on this matter.

sagulolo · ‎01-28-2019

Hi ChrisNZ, My apologies for late reply and thank for your guidance. Let me go through the article and test it first before i come back to you. Thank you so much for your time.

sagulolo · ‎01-24-2019

Hi SAS communities, I actually new to SAS code for trigger email, i just wanted to send email to my user report available to retrieve from share folder after my EG job completed, attached my code and screenshot:- filename mymail email 'myemail@yahoo.com.my' subject='report xxxx'; data _null_; file mymail; put 'Dear all,'; put ; put 'All reports available in shared folder.'; put '<html><body><table>'; PUT '<tr><td>Shared Folder</td><td>Report Name</td>'; PUT '<tr><td> test </td><td> test2 </td>'; PUT '</table></body></html>'; put ; put 'Thanks & Regards,'; run; Results: Expected: **link A actually is hyperlink to the share folder Please advice and guide me how to get the expected table, thanks you in advance.

sagulolo · ‎08-30-2018

Hi RW9, Understood and thank you, sir. Best Regards.

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