Invalid Procedure Call error

[sasasasa]

Hi,
I am getting the error saying 'Invalid Procedure Call' in my query. My query is very simple. I am just getting data from another query which is getting data from a table.

SELECT
Last(ProperCase (qryCustomerInf o.[Last Name]))
FROM qryCustomer

The qryCustomer is working fine and returning all the required rows but I don't have any idea why getting the last name is causing problem. Please help.

[Mariostg]

Hello.
You probably want something like this:

SELECT 
Last(StrConv(qryCustomerInfo.[Last Name]),3)
FROM qryCustomer

The 3 in the expression will render ProperCase, 2 would give lowercase and 1 uppercase.

[MMcCarthy]

Where did you get the ProperCase function from? Have you tried running the query without it.

[sasasasa]

If I use ProperCase(qryC ustomerInfo.[Last Name]), it works fine. If a customer's Last Name is ABC the it is converted to Abc. But when I add Last to the query it is causing problem. I think there is some problem in the data. But I don't know what can it be.

If I use Last(StrConv(qr yCustomerInfo.[Last Name],3)) then also it is saying Invalid procedure call. I don't know what is really happening. Please help me.

[Mariostg]

As far as I know, ProperCase does not exist, at least in Access 2003... That is why I proposed StrConv

In adidtion, you refer to qryCustomerInfo .[Last Name]
While you specify FROM qryCustomer... Something is not quite right here. I think in both it has to be the same.

[sasasasa]

I am sorry it is the same. It is qryCustomerInfo . Just a typing mistake here. My query is working fine if I get the data from a different table. I tried with StrConv but it is still displaying the same error.

[sasasasa]

I tried using only Last(qryCustome rInfo.[Last Name]) and it is still giving the same error. How can I find if there is something wrong in the data? Is there any other alternative of Last function?

[Mariostg]

I am getting clueless but can your run:

SELECT 
qryCustomerInfo.[Last Name]
FROM qryCustomerInfo

[sasasasa]

Yes, this runs fine.
My actual query is :
SELECT
qryCustomerInfo .[Number],
Last(qryCustome rInfo.[Last Name]) AS [LastOfLast Name], Last(qryCustome rInfo.[Phone])
FROM qryCustomerInfo
GROUP BY qryCustomerInfo .[Number]

Phone numbers are duplicate and I just want their last Phone numbers, so I am using this Last function.

I tried with only

SELECT
Last(qryCustome rInfo.[Last Name])
FROM qryCustomerInfo

it doesn't work. But it works if I remove

Last(qryCustome rInfo.[Last Name])

from my query. Phone numbers and Customer Numbers show up.

Your query worked just fine. But it has duplications. Last function is not working with Last Name.

[Mariostg]

Looks like you Select last of [last name] twice:
One here

SELECT 
qryCustomerInfo.[Number],
Last(qryCustomerInfo.[Last Name]) AS [LastOfLast Name], Last(qryCustomerInfo.[Phone])
FROM qryCustomerInfo
GROUP BY qryCustomerInfo.[Number]

and one here:

SELECT 
Last(ProperCase(qryCustomerInfo.[Last Name]))
FROM qryCustomerInfo

But qryCustomerInfo .[Last Name] does not exist.
qryCustomerInfo .[LastOfLast Name] exists on the other hand...
Not sure if you follow me.

So what you want is this

SELECT  
Last(ProperCase(qryCustomerInfo.[LastOfLast Name])) 
FROM qryCustomerInfo

[sasasasa]

I know I am confusing you. But actually I changed my query, as ProperCase was not a big issue for me, I removed it. I am using only

Last(qryCustome rInfo.[Last Name]) AS [LastOfLast Name]

now. So that I can make the normal query to work first and forget about Upper case or Lower case. So the query is just this

SELECT
qryCustomerInfo .[Number],
Last(qryCustome rInfo.[Last Name]) AS [LastOfLast Name], Last(qryCustome rInfo.[Phone])
FROM qryCustomerInfo
GROUP BY qryCustomerInfo .[Number]

[MMcCarthy]

Try something for me ...

SELECT qryCustomerInfo.[Number],
Last(qryCustomerInfo.[Last Name]) AS [LastOfLast Name], qryCustomerInfo.[Phone]
FROM qryCustomerInfo
GROUP BY qryCustomerInfo.[Number], qryCustomerInfo.[Phone]

Does this give you the result you want?

[sasasasa]

Yes, this works. Thank you sooooo much.

[MMcCarthy]

You're welcome :)