Feeder Question
- ioscat
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Feeder Question
Hello.
What may be the reason for not be working feeder:
[‘a’] => DB ( IF([]=0,’’,CubeName), … );
?
What may be the reason for not be working feeder:
[‘a’] => DB ( IF([]=0,’’,CubeName), … );
?
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Re: Feeder Question
Hello
What happens if you change the [ ] notation with the equivalent DB( ) notation?
Wim
What happens if you change the [ ] notation with the equivalent DB( ) notation?
Wim
Best regards,
Wim Gielis
IBM Champion 2024-2025
Excel Most Valuable Professional, 2011-2014
https://www.wimgielis.com ==> 121 TM1 articles and a lot of custom code
Newest blog article: Deleting elements quickly
Wim Gielis
IBM Champion 2024-2025
Excel Most Valuable Professional, 2011-2014
https://www.wimgielis.com ==> 121 TM1 articles and a lot of custom code
Newest blog article: Deleting elements quickly
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Re: Feeder Question
Why do so many posters insist on posting incomplete, hypothetical code, and then ask why it doesn't work. How do you expect us to help you when the code is incomplete? It could very well be a syntax problem (the cause of about 99% of user's problems) but we'll never know because you won't show us the REAL code. Firstly, why would you even write a conditional feeder like this, it makes no sense? By definition, a feeder only happens when the feeding cell is not zero, so show checking to see if it is zero first and then feeding only when not zero is not necessary. I can't help you beyond that because you didn't provide any code with which to determine your problem.ioscat wrote:Hello.
What may be the reason for not be working feeder:
[‘a’] => DB ( IF([]=0,’’,CubeName), … );
?
- ioscat
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Re: Feeder Question
i just had a try - no difference =(
well actually i'm trying to stop chain feedering from once overfed cells because we have task like:
Cube (dimA, dimB) ---> Cube (dimA, dimB, dimC) with some condition on dimB and dimC (some relations on matrix (dimB, dimC)) and the only way is to feed all elements in dimC. But most of the elements are zero since relation matrix is very sparsed.
well actually i'm trying to stop chain feedering from once overfed cells because we have task like:
Cube (dimA, dimB) ---> Cube (dimA, dimB, dimC) with some condition on dimB and dimC (some relations on matrix (dimB, dimC)) and the only way is to feed all elements in dimC. But most of the elements are zero since relation matrix is very sparsed.
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Re: Feeder Question
tomok,
the code is hypothetical because the question is hypothetical
(Identical cubes '1','2','3'; dims '123' and 'abc'; feeder from 1=>2, 2=>3; rule 1-->3)
the code is hypothetical because the question is hypothetical
Are you sure? I'm sure it's not.By definition, a feeder only happens when the feeding cell is not zero
(Identical cubes '1','2','3'; dims '123' and 'abc'; feeder from 1=>2, 2=>3; rule 1-->3)
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Re: Feeder Question
I'm a bit lost as to exactly what is trying to be done here but just one point of note:
You are trying to feed "Cube 3" from "Cube 2", "Cube 2" itself doesn't have SkipCheck enabled - this won't work.
Although it seems a bit odd since "Cube 2" itself has no rules, in order for feeders from "Cube 2" to another to work, you need to have SkipCheck enabled before your Feeders.
Hopefully that might help with one little issue, even if it doesn't deal with the whole thing. (Obviously also a cube needs to have values to be able to feed something else.)
You are trying to feed "Cube 3" from "Cube 2", "Cube 2" itself doesn't have SkipCheck enabled - this won't work.
Although it seems a bit odd since "Cube 2" itself has no rules, in order for feeders from "Cube 2" to another to work, you need to have SkipCheck enabled before your Feeders.
Hopefully that might help with one little issue, even if it doesn't deal with the whole thing. (Obviously also a cube needs to have values to be able to feed something else.)
Declan Rodger
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Re: Feeder Question
Your first post said absolutely nothing about a 3 cube structure where cube 1 feeds cube 2 which in turn feeds cube 3. As far as feeders is concerned, a fed cell and a non-zero cell are the same thing. If you are feeding from a cell that has been fed from somewhere else, that will also trigger a feeder. I didn't get that deep into the explanation because chaining feeders together like this is an exceedingly bad idea and should never really be done as it is almost impossible to debug. The bottom line is that the ORIGINAL cell that is triggering a feeder must be non-zero. In your example, the original cell is in Cube 1, and it is indeed not zero for the cells in question. What in the world that has to do with your original question escapes me. Also, if it is hypothetical, and you don't have any REAL CODE, how do you know it doesn't work??????
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Re: Feeder Question
tomok,
what do you mean at "chaining feeders together"?
what do you mean at "chaining feeders together"?
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Re: Feeder Question
I have to admit that I understand Tomok's frustration on this one.ioscat wrote:tomok,
the code is hypothetical because the question is hypothetical
I don't wish to discourage the free flow of "what if" conjecture, but if it is hypothetical you should at least be stating all, and that is ALL of the relevant information relating to the hypothesis. Such as the fact that three cubes are involved, for instance. Or whether you have in fact tried this "hypothetical" code. (As was pointed out in another post it's hard to see how you could know that it doesn't work if you hadn't tried it.) And what the real code is, not some abstract representation of it. Otherwise you're just wasting people's time, which nudges toward abusing the help on offer here.
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Re: Feeder Question
The example you posted is a perfect example. The original values, meaning actual data, is stored in Cube 1. You then have a feeder in Cube 1 that feeds the corresponding cells in Cube 2. This places a "marker" in those cells in Cube 2. However, since you don't have actually have a rule in Cube 2 to pull the values in, they never display. Regardless, they have been fed. If you click on one of these cells in Cube 2 and chose Check Feeders, you would see that they have been fed. Now, you have a feeder rule in Cube 2 that feeds Cube 3. The cells in Cube 2 that have been marked (fed from Cube 1), as well as any original data entered into Cube 2, will feed Cube 3. Since you actually have a rule in 3 that pulls in data from Cube 1, the values display AND have been fed, not because they were fed from Cube 1, but because they were fed to Cube 2, which then fed them to Cube 3. This is "chaining" of feeders. As you can see, it can be extremely confusing and should probably be avoided if at all possible.ioscat wrote:tomok,
what do you mean at "chaining feeders together"?