Hello,
I am facing problem with the list type variable
List1(1)=“A”
List1(1)=“B”
List1(2)=“C”
List1(3)=“D”
Now the problem with as List1 is having 2 same listtag so it is considering only 1 and displaying the final list as
List1(1)=“B”
List1(2)=“C”
List1(3)=“D”
Rather, I am trying for
List1(1)=“A”
List1(1)=“B”
List1(2)=“C”
List1(3)=“D”
Please suggest me regarding this
Thanks
Subject: Try this for example
dim list1 list as variant
dim firstListEntry(1) as variant
dim secondListEntry(1) as variant
firstListEntry(0)=“A”
firstListEntry(1)=“B”
secondListEntry(0)=“A”
secondListEntry(1)=“C”
list1(“1”)=firstListEntry
list1(“2”)=secondListEntry
Subject: List Problem
You can’t - what do you want to get returned if you look at the value of List(1)? You can’t have an array within a list.
You’d need to concatenate the two, with a delimiter which you can then parse. I tend to use the ~
So when setting the values
if iselement(list1(1)) then
list1(1) = list(1) + “~B”
Else
list(1) = “A”
End If
At least that’s what I assume you want to do, you don’t give much info
Subject: RE: List Problem
I have RequestInfo List as Variant
Now RequestInfo contains some information on every listtag.
RequestInfo(RequestNo) = Values
Now I do not know the RequestNo as it is fetched from some other view.
and if RequestNo comes with duplicate entry then it takes one out of 2 request nos.
I want to display both RequestNo.
Thanks
Subject: RE: List Problem
OK, well I just tested this and you can apply a variant array to a list. But you will need to create your array separately. As a simple example:
dim RequestInfo list as variant
dim varray as variant
redim varray(1)
varray(0) = “A”
varray(1) = “B”
RequestInfo(1) = varray
If you reassign a new item to RequestInfo it does write over it (as you’ve found), so you’ll have to do something like the above if you want to use arrays
Subject: RE: List Problem
I do not want to use array.
I have only a list i.e. caseInfo(CaseNo) = CaseDetails
Now if CaseNo is duplicate say
CaseNo = “2131”
CaseNo = “2323”
CaseNo = “2131”
When these case nos. insert in
caseInfo(CaseNo) = CaseDetails
It takes “2131” CaseNo only one time so insipte of displaying the values of
caseInfo(2131) = CaseDetails
caseInfo(2323) = CaseDetails
caseInfo(2131) = CaseDetails,
it is displaying the values of
caseInfo(2323) = CaseDetails
caseInfo(2131) = CaseDetails
Subject: The problem is not the list
The list is working the way it is designed, the problem is you are not understanding how a list works.
You have the example:
List1(1)=“A”
List1(1)=“B”
List1(2)=“C”
List1(3)=“D”
You’re thinking that because you wrote it that way it’s stored that way, it isn’t. When you put you second List1(1)=“B” the original List1(1)=“A” value is gone, finis, none existent.
If you think of it this way I think you’ll realise why what you’re asking for doesn’t work the way you think it should.
Val1 = A
Val1 = B
Val2 = C
Val3 = D
Subject: RE: List Problem
That’s the way it works. I’ve given you a couple of suggestions as to how to get round it, there’s no point in telling me again what you’re doing. If what you’re doing isn’t possible you have to find another way of doing it…