Dictionary Object

[Johnny Klunk]

Wondering if someone can give me a hand with something that I'm sure is
really easy - but damned if I know what I'm doing wrong. I'm trying to read
the contents of a database into an ASP dictionary object. However I'm
getting the error

Microsoft VBScript runtime error '800a01c9'
This key is already associated with an element of this collection
/test.asp, line 21

There's definately no repeated data in either column, it's currently only
test data, one column is 1,2,3,4 etc the other is a,b,c,d. I can confirm
it's reading the data properly if I use response.write instead of d.add
The code I'm using is this:

Set rsCount = Server.CreateOb ject ("ADODB.Records et")
sqlView = "select data,number from tblTemp"
Set d=Server.Create Object("Scripti ng.Dictionary")

rsCount.Open sqlView, "dsn=dsn_hits_2 "
do while not rsCount.eof
d.add rsCount("number "),rsCount("dat a")
rsCount.MoveNex t
loop
rsCount.close

Set rsCount = Nothing




Many thanks in advance.

[Ray at]

Re: Dictionary Object

When you add an item to the dictionary object, the key must be unique. The
key is the first argument in the add method, which in this case, is
rsCount("number "). What that error means is that it's hitting a record that
has rsCount("number ") with the same value as another record that has already
been inserted into the dictionary object. i.e.


number data
1 Joe
2 Kelly
3 Frank
2 Jorg

When it gets to the Jorg record and tries to use 2 as the key, that'll throw
an error since the Kelly item in the dictionary object already has a key of
2.

Possible Solutions:

Use an array instead of a dictionary object.

Use the primary key in your recordset if the key doesn't matter.

Use a counter for the key and increase it by one in your loop if the key
doesn't matter.

Probably some other things.

Ray at work


"Johnny Klunk" <johnnyklunk@:r em0ve-this:johnnyklun k.com> wrote in message
news:befbpm$47q $1@sparta.btint ernet.com...[color=blue]
> Wondering if someone can give me a hand with something that I'm sure is
> really easy - but damned if I know what I'm doing wrong. I'm trying to[/color]
read[color=blue]
> the contents of a database into an ASP dictionary object. However I'm
> getting the error
>
> Microsoft VBScript runtime error '800a01c9'
> This key is already associated with an element of this collection
> /test.asp, line 21
>
> There's definately no repeated data in either column, it's currently only
> test data, one column is 1,2,3,4 etc the other is a,b,c,d. I can confirm
> it's reading the data properly if I use response.write instead of d.add
> The code I'm using is this:
>
> Set rsCount = Server.CreateOb ject ("ADODB.Records et")
> sqlView = "select data,number from tblTemp"
> Set d=Server.Create Object("Scripti ng.Dictionary")
>
> rsCount.Open sqlView, "dsn=dsn_hits_2 "
> do while not rsCount.eof
> d.add rsCount("number "),rsCount("dat a")
> rsCount.MoveNex t
> loop
> rsCount.close
>
> Set rsCount = Nothing
>
>
>
>
> Many thanks in advance.
>
>[/color]

[Johnny Klunk]

Re: Dictionary Object


"Ray at <%=sLocation% >" <ask@me.forit > wrote in message
news:O3txsYZRDH A.2248@TK2MSFTN GP12.phx.gbl...[color=blue]
> When you add an item to the dictionary object, the key must be unique.[/color]
The[color=blue]
> key is the first argument in the add method, which in this case, is
> rsCount("number "). What that error means is that it's hitting a record[/color]
that[color=blue]
> has rsCount("number ") with the same value as another record that has[/color]
already[color=blue]
> been inserted into the dictionary object. i.e.[/color]


Hi,
Thanks for the response. The thing is, the data is definately not repeated.
I entered the data manually to test and make there's no error. It's just
1,2,3,4 etc..
That's a good idea with the primary key. I've just tested again, setting
each of my columns as the primary key to ensure the database has no repeated
entries. I'm still getting the same thing. It's driving me nuts!

[Ray at]

Re: Dictionary Object

Try this then and look:

rsCount.Open sqlView, "dsn=dsn_hits_2 "
do while not rsCount.eof
response.write rsCount("number ") & "," & rsCount("data") &
"<br />"
rsCount.MoveNex t
loop
rsCount.close

Ray at work


"Johnny Klunk" <johnnyklunk@:r em0ve-this:johnnyklun k.com> wrote in message
news:befd6c$rv5 $1@hercules.bti nternet.com...[color=blue]
>
> "Ray at <%=sLocation% >" <ask@me.forit > wrote in message
> news:O3txsYZRDH A.2248@TK2MSFTN GP12.phx.gbl...[color=green]
> > When you add an item to the dictionary object, the key must be unique.[/color]
> The[color=green]
> > key is the first argument in the add method, which in this case, is
> > rsCount("number "). What that error means is that it's hitting a record[/color]
> that[color=green]
> > has rsCount("number ") with the same value as another record that has[/color]
> already[color=green]
> > been inserted into the dictionary object. i.e.[/color]
>
>
> Hi,
> Thanks for the response. The thing is, the data is definately not[/color]
repeated.[color=blue]
> I entered the data manually to test and make there's no error. It's just
> 1,2,3,4 etc..
> That's a good idea with the primary key. I've just tested again, setting
> each of my columns as the primary key to ensure the database has no[/color]
repeated[color=blue]
> entries. I'm still getting the same thing. It's driving me nuts!
>
>[/color]

[Johnny Klunk]

Re: Dictionary Object


"Ray at <%=sLocation% >" <ask@me.forit > wrote in message
news:e7IwBlZRDH A.1924@TK2MSFTN GP12.phx.gbl...[color=blue]
> Try this then and look:
>
> rsCount.Open sqlView, "dsn=dsn_hits_2 "
> do while not rsCount.eof
> response.write rsCount("number ") & "," & rsCount("data") &
> "<br />"
> rsCount.MoveNex t
> loop
> rsCount.close
>
> Ray at work
>[/color]

Ray,
Thanks again for the quick response :) I've actually tried that idea - but
to be sure, to be sure I pasted your code in to make sure the results
matched. I've copy-pasted the output from test.asp below. You can see why
this is driving me nuts ! As far as I see, every row has different data in
each column.

a,1
b,2
c,3
d,4
e,5
f,6
g,7

[Chris Hohmann]

Re: Dictionary Object

"Johnny Klunk" <johnnyklunk@:r em0ve-this:johnnyklun k.com> wrote in
message news:befbpm$47q $1@sparta.btint ernet.com...
[color=blue]
>d.add rsCount("number "),rsCount("dat a")[/color]

Try:
d.add rsCount.Fields. Item("number"). Value,
rsCount.Fields. Item("data").Va lue

Also you can take advantage of the fact that the dictionary object will
create/overwrite a key based on whether it exists or not:
d(rsCount.Field s.Item("number" ).Value) =
rsCount.Fields. Item("data").Va lue

However this method wreaks of "programmin g by side effect" so if the
first method work, use that.

HTH
-Chris

[Chris Hohmann]

Re: Dictionary Object

"Johnny Klunk" <johnnyklunk@:r em0ve-this:johnnyklun k.com> wrote in
message news:befga2$cga $1@sparta.btint ernet.com...[color=blue]
>[color=green]
> > d.add rsCount.Fields. Item("number"). Value,[/color]
> rsCount.Fields. Item("data").Va lue[color=green]
> >[/color]
>
>
> Yep, thats worked. Superb, thanks so much. No idea why my bit of[/color]
code[color=blue]
> didn't work, but I'm glad for a solution.
>
> Cheers
>
>[/color]
You code was attempting to set the dictionary key to a field Object, not
it's value. Whenever possible, try to explicitly reference values. When
you reference the value of a field object implicitly (i.e.
rsCount("number ") you're asking the parser to do the following:

1. Determine the default property/method for the rsCount object, the
Fields collection
2. Determine the default property/method for the Fields collection, the
Item method
3. Determine the default property/method for the Item object, the Value
property
4. Determine if the Value property is an object
5. If it is an object do the whole object model default method/property
traversal thing again.
6. If not, return the Value

All of these steps are also dependent on the context in which they are
called. So sometimes the default thing happens, sometime it doesn't.
There are also a lot more going on before, between and after each of
these steps, but you get the idea. Always better to explicitly identify
the "value" you want.

HTH-
Chris