Jim DeMarco
Jdemarco at hshhp.org
Fri Dec 19 07:35:31 CST 2003
Doh!!!
Thanks,
Jim D
-----Original Message-----
From: A.D.Tejpal [mailto:adtp at touchtelindia.net]
Sent: Thursday, December 18, 2003 11:39 PM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] The ARRAY function
Jim (DeMarco),
My communication was intended for Jim (Lawrence) and meant to reinforce your excellent advice to him.
Regards,
A.D.Tejpal
--------------
----- Original Message -----
From: Jim DeMarco
To: Access Developers discussion and problem solving
Sent: Friday, December 19, 2003 01:50
Subject: RE: [AccessD] The ARRAY function
I think you misunderstood me. Yes you can pass variables to Array as arguments as long as you separate your arguments with comma (as you've done here). I was pointing out that the string "Tom, Dick, Harry" passed into Array as one string is not the same thing as passing each individual item as separate arguments to Array (whether literal or variable).
Jim DeMarco
-----Original Message-----
From: A.D.Tejpal [mailto:adtp at touchtelindia.net]
Sent: Thursday, December 18, 2003 1:03 PM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] The ARRAY function
Jim,
Basically, there is nothing wrong with the use of variables as arguments to an Array() function. However, it has to be kept in mind that such arguments have to be interpretable as comma-delimited list of values.
In the code originally sent by you, the whole string represented by variable StrStuff was getting interpreted as one vale (the commas treated as internal to the string). As a result you were getting a single element array and the value of varArray(0) was correctly being shown as the string containing all the three names.
Sample code given below illustrates the use of variables as arguments to Array() function.
Regards,
A.D.Tejpal
--------------
-------------
Code Start
-------------
Sub TestArray()
' Return value of Array function is Variant type
Dim varArray As Variant, StrStuff1 As String
Dim StrStuff2 As String, StrStuff3 As String
StrStuff1 = "Tom"
StrStuff2 = "Dick"
StrStuff3 = "Harry"
' Print strStuff1 To 3
Debug.Print StrStuff1 & ", " & StrStuff2 & ", " & StrStuff3
' Output = Tom, Dick, Harry
' Run the ARRAY function to place the contents of
' strings into an array. (arguments for Array function
' have to be in the form of a comma-delimited list of values)
varArray = Array(StrStuff1, StrStuff2, StrStuff3)
' Print the first index.
Debug.Print varArray(0)
' Output = Tom
End Sub
-------------
Code End
-------------
----- Original Message -----
From: Jim Lawrence (AccessD)
To: Access Developers discussion and problem solving
Sent: Thursday, December 18, 2003 21:23
Subject: RE: [AccessD] The ARRAY function
Jim
Thanks for the information...much appreciated.
The actual line feed in through variable was:
"Tom", "Dick", "Harry" not "Tom, Dick, Harry"
Visually there is no difference between the manually entered string and the
'generated' variable.
I think the split function will work.
Again THanks
Jim
-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Jim DeMarco
Sent: Thursday, December 18, 2003 5:16 AM
To: Access Developers discussion and problem solving
Subject: RE: [AccessD] The ARRAY function
Yes that solved your problem but maybe some explanation on why your original
code failed might help others? If not please read on.
The difference between:
strStuff = "Tom, Dick, Harry"
varArray = Array(strStuff)
Debug.Print varArray(0) 'prints "Tom, Dick, Harry"
and
varArray = Array ("Tom", "Dick", "Harry")
Debug.Print varArray(0) 'prints "Tom"
is that the Array function needs to be fed the individual array elements as
separate arguments. Yes they look comma delimited but it's just to separate
the arguments. Feeding in variable that contains strings separated by a
comma is not the same thing.
The Split function that worked for you took your string a created an array
based on the "," delimiter which was then assigned to your variant.
Jim DeMarco
-----Original Message-----
From: Jim Lawrence (AccessD) [mailto:accessd at shaw.ca]
Sent: Wednesday, December 17, 2003 9:39 PM
To: Access Developers discussion and problem solving
Subject: RE: [AccessD] The ARRAY function
Brilliant; Charlotte
Many thanks.
Jim
-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Charlotte
Foust
Sent: Wednesday, December 17, 2003 5:57 PM
To: Access Developers discussion and problem solving
Subject: RE: [AccessD] The ARRAY function
If you want to use a variable, use Split instead of Array(). Then you
don't need all those extra chr(34)s, etc.
Public Function TestArray()
Dim strStuff As String
Dim vararray As Variant
strStuff = "Tom, Dick, Harry"
Debug.Print strStuff
' Run the Split function to place the contents of the string into an
array.
vararray = Split(strStuff, ",")
Debug.Print vararray(0)
End Function
Charlotte Foust
-----Original Message-----
From: Jim Lawrence (AccessD) [mailto:accessd at shaw.ca]
Sent: Wednesday, December 17, 2003 5:28 PM
To: Access Developers discussion and problem solving
Subject: [AccessD] The ARRAY function
Hi All:
I am running in some strange responses, maybe too much chocolate has
fuddled the brain but:
Given the following:
' Set strStuff
strStuff = chr(34) & "Tom" & chr(34) & ", " & chr(34) & "Dick" & chr(34)
& ", " & chr(34) & "Harry" & chr(34) ' Print strStuff ? strStuff "Tom",
"Dick", "Harry" ' Run the ARRAY function to place the contents of the
string into an array. varArray = Array(strStuff) ? varArray(0) ' Print
the first index. "Tom", "Dick", "Harry" ' OK ????? ' Now type in the
same information manually or cut and paste... varArray = Array("Tom",
"Dick", "Harry") ? varArray(0) ' and the results are what they should
be. Tom
What is going on and why can a variable not be used in the ARRAY
function?? :-(
Any suggestions would be greatly appreciated.
Jim
_______________________________________________
AccessD mailing list
AccessD at databaseadvisors.com
http://databaseadvisors.com/mailman/listinfo/accessd
Website: http://www.databaseadvisors.com
***********************************************************************************
"This electronic message is intended to be for the use only of the named recipient, and may contain information from Hudson Health Plan (HHP) that is confidential or privileged. If you are not the intended recipient, you are hereby notified that any disclosure, copying, distribution or use of the contents of this message is strictly prohibited. If you have received this message in error or are not the named recipient, please notify us immediately, either by contacting the sender at the electronic mail address noted above or calling HHP at (914) 631-1611. If you are not the intended recipient, please do not forward this email to anyone, and delete and destroy all copies of this message. Thank You".
***********************************************************************************