The question of using Set analysis in Dimension

[#1]

I have a testing table as below.

Test:

Load * inline

[id,amt

1,10

1,20

1,30

2,10

2,20

2,30];

I would like to show all the IDs in the object always, although the ID is filtered in the selection.

I created a straight table as following, but it not work. it only show the id which filter by ID in the selection, but I would like to show all the IDs.

dimension:

=aggr(MaxString({1<id=>} id),id)

expression:

sum(amt)

could you please help me?

thanks a lot.

[Gysbert_Wassenaar]

You can add an expression sum({<id=>}1) to the straight table. You can then hide that expression. See attached example.

[swuehl]

Have you tried using set analysis in the expression?

id

as dimension

sum({1} amt)

as expression

[Anonymous]

You can use id as dimension, not a calculated dimension - and check "show all values".

But this is not enough.  You have also take care of expressions.  Try to uncheck "suppress zero values" and "supress missing".

Regards,

Michael

[#5]

Thanks for your help guys, but I think I did not explain chear. the expreesions are so complex in my actual case, so your suggestions are not suit to my situation.

I confused that why the Set analysis is not work in the Dimension.

for my exsample below, if I would like to show ID 1 and 2 always, how to set? Thanks.

Test:

Load * inline

[id,amt

1,10

1,20

1,30

2,10

2,20

2,30

3,10

3,20];

[kaushiknsolanki]

Hi,

     Create a straight table with dimension as ID

     and expression as

     Sum({1<ID = {"1","2"}>}amt)

Regards,

Kaushik Solanki