Hey Everybody,
I have code for a command button. The button is supposed to open a query with a general criteria of: Forms![Form Name]![Control Name]. The [Column Name] field of the query should match the [Control Name] in the Form. If a record does not exist in the query for the specified control name, I want a message box to display stating so. My code is below: - If ( ) Then ' If record exists, then open the query below.
-
DoCmd.OpenQuery "2 - Total Reflux Tests Query", acViewNormal, acReadOnly
-
Else ' If record does not exist, then show message below.
-
MsgBox "No test(s) found for column " & [Column Name] & ".", vbInformation, "Sorry"
-
End If
What should go in the if statement?
Thanks!
15 2004
Hey Everybody,
I have code for a command button. The button is supposed to open a query with a general criteria of: Forms![Form Name]![Control Name]. The [Column Name] field of the query should match the [Control Name] in the Form. If a record does not exist in the query for the specified control name, I want a message box to display stating so. My code is below: - If ( ) Then ' If record exists, then open the query below.
-
DoCmd.OpenQuery "2 - Total Reflux Tests Query", acViewNormal, acReadOnly
-
Else ' If record does not exist, then show message below.
-
MsgBox "No test(s) found for column " & [Column Name] & ".", vbInformation, "Sorry"
-
End If
What should go in the if statement?
Thanks!
I've tried this expression in the If statement: - [2 - Total Reflux Tests]![Column Name] = Forms![GeneralColumnSummary - view]![Column Name]
But an error message popped up when I tried to run it:
"Run-time error '2465':
Microsoft Office Access can't find the field '|' referred to in your expression."
What can I do to fix it?
Thanks!
I've tried this expression in the If statement: - [2 - Total Reflux Tests]![Column Name] = Forms![GeneralColumnSummary - view]![Column Name]
But an error message popped up when I tried to run it:
"Run-time error '2465':
Microsoft Office Access can't find the field '|' referred to in your expression."
What can I do to fix it?
Thanks!
Can someone please help me???
NeoPa 32,568
Recognized Expert Moderator MVP
Please do not *BUMP* your thread before at least 24 hours have passed.
Our experts and members offer their time voluntarily and do not deserve to be treated with impatience by those who may benefit from their efforts.
MODERATOR.
I will look to answer this later but, like all the other experts, I don't appreciate your impatience. Having received a PM from you too, I'm not feeling too generous towards you at the moment. I'll look at this when I get home from work.
Please do not *BUMP* your thread before at least 24 hours have passed.
Our experts and members offer their time voluntarily and do not deserve to be treated with impatience by those who may benefit from their efforts.
MODERATOR.
I will look to answer this later but, like all the other experts, I don't appreciate your impatience. Having received a PM from you too, I'm not feeling too generous towards you at the moment. I'll look at this when I get home from work.
To NeoPa and all other Moderators:
I am terribly sorry for coming accross as impatient. I am new to writing on Forums (I didn't even know what "BUMP" means), and thus I had no clue that what I was doing was considered impatient. I also did not know that sending you personal messages requesting help was frowned upon. So, to prevent anymore mistakes and annoyances, I will resign myself from this forum. I'm sorry for upsetting all of you. I really, truly am sorry.
~mforema
JKing 1,206
Recognized Expert Top Contributor
Adding this might work.
The logic behind this is to open the query and search for the first record that matches your control. If there is an existing record with the specified criteria Nomatch will return false. So, in theory this should provide you with the correct logic for your If statement.
Dim db as Database
Dim rstQuery As RecordSet
Set db = CurrentDb
Set rstQuery = db.OpenRecordSe t ("qryName", dpOpenDynaset)
rstQuery.FindFi rst ("[Column Name] = [Forms]![Form]![Control Name]")
If rstQuery.NoMatc h = False Then
*the rest of your code here*
Hope this was helpful.
Jking
Adding this might work.
The logic behind this is to open the query and search for the first record that matches your control. If there is an existing record with the specified criteria Nomatch will return false. So, in theory this should provide you with the correct logic for your If statement.
Dim db as Database
Dim rstQuery As RecordSet
Set db = CurrentDb
Set rstQuery = db.OpenRecordSe t ("qryName", dpOpenDynaset)
rstQuery.FindFi rst ("[Column Name] = [Forms]![Form]![Control Name]")
If rstQuery.NoMatc h = False Then
*the rest of your code here*
Hope this was helpful.
Jking
Thanks JKing
I tried your code, but the an error message popped up saying "Too few parameters, expected:1." I don't know what that means. When trying to debug, the "Set rstQuery" line was highlighted. Here is my code so far: - Dim db As Database
-
Dim rstQuery As Recordset
-
-
Set db = CurrentDb
-
Set rstQuery = db.OpenRecordset("2 - Total Reflux Tests Query", dbOpenDynaset)
-
-
rstQuery.FindFirst ("[Column Name] = [Forms]![GeneralColumnSummary - view]![Control Name]")
-
-
If (rstQuery.NoMatch = False) Then
-
DoCmd.OpenQuery "2 - Total Reflux Tests Query", acViewNormal, acReadOnly
-
Else
-
MsgBox "No tests found for column " & [Column Name] & ".", vbInformation, "Sorry"
-
End If
Do you see what I did wrong?
Thanks again.
NeoPa 32,568
Recognized Expert Moderator MVP
To NeoPa and all other Moderators:
I am terribly sorry for coming accross as impatient. I am new to writing on Forums (I didn't even know what "BUMP" means), and thus I had no clue that what I was doing was considered impatient. I also did not know that sending you personal messages requesting help was frowned upon. So, to prevent anymore mistakes and annoyances, I will resign myself from this forum. I'm sorry for upsetting all of you. I really, truly am sorry.
~mforema
We're really not all that thin-skinned, don't worry.
Simply, we have to deal with very large numbers of people of all sorts here, so replying in a way that is right for everyone is just not possible. I wanted to impart to you the understanding (which seems to have got across quite adequately) that such things are frowned on (and why). In my reply PM I posted a link so that you can better understand what we're dealing with here, and how members should be using this free resource that is TSDN.
I have no wish for you to leave the forums - that is not what we're all about.
Please enjoy the rest of your stay here.
NeoPa 32,568
Recognized Expert Moderator MVP
Thanks JKing
I tried your code, but the an error message popped up saying "Too few parameters, expected:1." I don't know what that means. When trying to debug, the "Set rstQuery" line was highlighted. Here is my code so far: - Dim db As Database
-
Dim rstQuery As Recordset
-
-
Set db = CurrentDb
-
Set rstQuery = db.OpenRecordset("2 - Total Reflux Tests Query", dbOpenDynaset)
-
-
rstQuery.FindFirst ("[Column Name] = [Forms]![GeneralColumnSummary - view]![Control Name]")
-
-
If (rstQuery.NoMatch = False) Then
-
DoCmd.OpenQuery "2 - Total Reflux Tests Query", acViewNormal, acReadOnly
-
Else
-
MsgBox "No tests found for column " & [Column Name] & ".", vbInformation, "Sorry"
-
End If
Do you see what I did wrong?
Thanks again.
I've checked the line in question and I can't see what your problem could be.
Have you copied /pasted this code in. Sometimes it is very important that you do (Always do so as a general rule).
Is the query name absolutely accurate, and does the query run when opened from the database window?
Can you post the SQL of the query for us to look at please.
NeoPa 32,568
Recognized Expert Moderator MVP
I've tried this expression in the If statement: - [2 - Total Reflux Tests]![Column Name] = Forms![GeneralColumnSummary - view]![Column Name]
But an error message popped up when I tried to run it:
"Run-time error '2465':
Microsoft Office Access can't find the field '|' referred to in your expression."
What can I do to fix it?
Thanks!
Going back to your earlier post, can you explain in simple English what the two sides of the = test refer to.
Again, nothing jumps out as to why the compiler would think there were an '|' in the equation anywhere.
PS. Lots of traffic since your last post. Best if you review all of it in sequence.
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