Subject | Re: [IBO] Need fastest record count for condition |
---|---|
Author | James N Hitz |
Post date | 2008-02-15T13:42:17Z |
Try this:
SELECT parentField1,
parentField2,
parentFieldN,
(select count(*) from ChildTable
where ChildTable.ForeignKey = ParentTable.PrimaryKey)
AS ChildCount
FROM parentTable
...and then you can just read IBOQuery.FieldByName('ChildCount').AsInteger
James
ifitsx wrote:
SELECT parentField1,
parentField2,
parentFieldN,
(select count(*) from ChildTable
where ChildTable.ForeignKey = ParentTable.PrimaryKey)
AS ChildCount
FROM parentTable
...and then you can just read IBOQuery.FieldByName('ChildCount').AsInteger
James
ifitsx wrote:
>
> w/IBO 4.2Hd, D7, and Fb 1.54
>
> I'm working with a VirtualTreeview and two Tables:
>
> One Table contains data on the folder nodes for the
> VirtualTreeview, including whether the folder has
> any documents that it owns.
>
> The other Table contains rows for documents - notes -
> including a blob.
>
> The VirtualTreeview must know the exact count of
> child nodes it will have, when the parent node is created.
>
> This is currently maintained by a select statement on
> the notes table with the simple criteria for the folder parent node,
> and then using the IBOQuery.RecordCount value passed to the
> VirtualTreeview InitChildren event.
> <http://groups.yahoo.com/group/IBObjects/message/42693;_ylc=X3oDMTM1Y2twbWFhBF9TAzk3MzU5NzE0BGdycElkAzQwMjkzMARncnBzcElkAzE3MDUwMDcxODMEbXNnSWQDNDI2OTMEc2VjA2Z0cgRzbGsDdnRwYwRzdGltZQMxMjAzMDMzMjM2BHRwY0lkAzQyNjkz>
>
> <http://groups.yahoo.com/;_ylc=X3oDMTJjOGpmbm1iBF9TAzk3MzU5NzE0BGdycElkAzQwMjkzMARncnBzcElkAzE3MDUwMDcxODMEc2VjA2Z0cgRzbGsDZ2ZwBHN0aW1lAzEyMDMwMzMyMzY->
>
>
> <http://groups.yahoo.com/group/IBObjects;_ylc=X3oDMTJkYm05dmtwBF9TAzk3MzU5NzE0BGdycElkAzQwMjkzMARncnBzcElkAzE3MDUwMDcxODMEc2VjA3Z0bARzbGsDdmdocARzdGltZQMxMjAzMDMzMjM2>
>
>
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