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How to use Feature_ID in Relation for suppress/Exclude single feature?

jtrivedi
1-Newbie

How to use Feature_ID in Relation for suppress/Exclude single feature?

How to use Feature_ID in Relation for suppress/Exclude single feature?

Please Help if anyone having experience on it.

14 REPLIES 14
MartinHanak
23-Emerald V
(To:jtrivedi)

Hi,

you cannot suppress feature using relations. You have to use Pro/PROGRAM functionality.

MH


Martin Hanák

Dear Martin,

I knew it and also i use from that side but i want to do using relation.

MartinHanak
23-Emerald V
(To:jtrivedi)

Hi,

I am sorry I do not understand what do you want to know.

MH


Martin Hanák

i use pro-program functionality for suppress feature at particular relation.

But i want to suppress feature with the use of Relation and Parameter.

MartinHanak
23-Emerald V
(To:jtrivedi)

Hi,

please provide detailed example explaining your need. For example: Attach model and describe, how do you want to ""drive" it. Use combination of  pictures and text when providing the explanation.

MH


Martin Hanák

‌I think they want to set up an IF-ELSE-ENDIF section in Pro/Program that controls whether a feature or features are suppressed or unsuppressed based on the value of a parameter.

Kevin

You are right.

I want to use IF Endif formula for suppress feature but i want to use by Relation and Parameter not using by Pro-Program.

TomU
23-Emerald II
(To:jtrivedi)

The only way to suppress features is with Pro/Program.   All the of the decision making logic can be in relations and parameters, but the actual suppress and resume can only be performed by Pro/Program.

ss3
5-Regular Member
5-Regular Member
(To:TomU)

please can suggest how to create parameters, and how to write a pro program to suppress a part.

Please got through complete below example how am i expecting.

 

Example: Consider XXX.asm is my master assembly. XXX-A is sub-assembly & is having n number of parts, i need to write a program such that among n number of parts, only 3 sheetmetal parts should be controlled.(SHT1, SHT2, SHT3)

 

That is, if WIDTH of my master assembly is (WIDTH<=1000MM & WIDTH>=850MM), SHT1.PRT Should resume. SHT2 and SHT3 should get suppress.

If WIDTH of my master assembly is greater than 1000mm(WIDTH>=1001 & WIDTH<=1600), SHT2.PRT & SHT3.PRT should get resume and SHT1.PRT should get suppress. (i dont want to use pattern or family table as SHT1/2/3 are of different sizes)

 

can Width be defined in increment or decrement value of 25mm? how the program should be?

 

Below are the internal component id:

ADD PART SHT1

INTERNAL COMPONENT ID 48

END ADD

 

ADD PART SHT2

INTERNAL COMPONENT ID 49

END ADD

 

ADD PART SHT3

INTERNAL COMPONENT ID 52

END ADD

 

I TRIED THIS: (BUT GETTING ERROR)

 

IF WIDTH=1200
errorLogical expression expected.
ADD SUPPRESSED PART SHT2.PRT
errorExtra symbols found - ignored.
INTERNAL COMPONENT ID 49
errorExtra symbols found - ignored.
ENDIF

 

THANK YOU

HamsterNL
17-Peridot
(To:ss3)

IF WIDTH==1200

ADD PART SHT2

INTERNAL COMPONENT ID 49

END ADD

END IF

 

Notice the "double equal sign" in the IF statement. If you are comparing, you will need ==

dschenken
21-Topaz I
(To:ss3)

Try to ask on one forum at a time. Some of us watch several of them and it's a time waster to respond on one while someone else responds on another. You will find that the speed of response is not much quicker but people tire of extra work and will slow their response. Also remember to go to all the forums you left the question on when you have an answer and post the answer.

 

(and it would be so much better if this forum software indicated the name and time the response was to.) 

ss3
5-Regular Member
5-Regular Member
(To:dschenken)

@dschenken It was a mistake, i was searching for solutions in other forums as well, and it was similar kind for me, so posted here, But to continue my thread posted there as well, so that it can help similar kind of problem facing people.

 

Thank you for your precious time.

ranjith_12
3-Visitor
(To:ss3)

Similar case i am also getting Error msg. Can any one help on this 

 

IF SHELF == "NO"
ADD SUPPRESSED SUBASSEMBLY AUTOCRATE96X48_150LBS_A05
INTERNAL COMPONENT ID 447
PARENTS = 80(#12) 68(#5)
END ADD
ELSE
ADD SUBASSEMBLY AUTOCRATE96X48_150LBS_A05
INTERNAL COMPONENT ID 447
PARENTS = 80(#12) 68(#5)
END ADD
!*** ERR: Feature with ID 447 already exists in the model.
END IF

IF SHELF == "YES"
ADD SUPPRESSED PART AUTOCRATE96X48_150LBS_P39
INTERNAL COMPONENT ID 453
PARENTS = 80(#12) 114(#26)
END ADD
ELSE
ADD PART AUTOCRATE96X48_150LBS_P39
INTERNAL COMPONENT ID 453
PARENTS = 80(#12) 114(#26)
END IF

!*** ERR: no END ADD statement for ADD on line 0
!*** ERR: file contains more IF's than ENDIF's

Just checking: if SHELF is a STRING, then you will need to use quotes ("). If SHELF is a YES_NO (boolean), then you'll only need to use YES and NO (without quotes)

 

IF SHELF == "YES"
   ADD PART AUTOCRATE96X48_150LBS_P39
   INTERNAL COMPONENT ID 453
   PARENTS = 80(#12) 114(#26) 
   END ADD

ELSE

   ADD SUBASSEMBLY AUTOCRATE96X48_150LBS_A05
   INTERNAL COMPONENT ID 447
   PARENTS = 80(#12) 68(#5) 
   END ADD
END IF

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