I'm using the if statement to check whether a cell is a number. If yes, just return the original number, else, return the string 'Not a number'. However, I got some troubles while I was outputting the string 'Not a number'. Following are the function I'm using, how could I modify it?
=if(isnumber(A1), A1, 'Not a number')
It works while A1 is a number but fails while it's not a number.
You are missing double quotes.
Try this
=if(isnumber(A1), A1,"Not a number")
Related
I'm getting this message in Redshift: invalid input syntax for type numeric: " " , even after trying to implement the advice found in SO.
I am trying to convert text to number.
In my inner join, I try to make sure that the text being processed is first converted to null when there is an empty string, like so:
nullif(trim(atl.original_pricev::text),'') as original_price
... I noticed from a related post on coalesce that you have to convert the value to text before you can try and nullif it.
Then in the outer join, I test to see that there's a limited set of acceptable characters and if this test is met I try to do the to_number conversion:
,case
when regexp_instr(trim(atl.original_price),'[^0-9.$,]')=0
then to_number(atl.original_price,'FM999999999D00')
else null
end as original_price2
At this point I get the above error and unfortunately I can't see the details in datagrip to get the offending value.
So my questions are:
I notice that there is an empty space in my error message:
invalid input syntax for type numeric: " " . Does this error have the exact same meaning as
invalid input syntax for type numeric:'' which is what I see in similar posts??
Of course: what am I doing wrong?
Thanks!
It's hard to know for sure without some data and the complete code to try and reproduce the example, but as some have mentioned in the comments the most likely cause is the to_number() function you are using.
In the earlier code fragment you are converting original_price to text (string) and then substituting an empty string ('') if the value is NULL. Calling the to_number() function on an empty string will give you the error described.
Without the full SQL statement it's not clear why you're putting the nullif() function around the original_price in the "inner join" or how whether the CASE statement is really in an outer join clause or one of the columns returned by the query. However you could perhaps alter the nullif() to substitute a value that can be converted to a number e.g. '0.00' instead of ''.
Sorry I couldn't share real data. I spent the weekend testing small sets to try and trap the error. I found that the error was caused by the input string having no numbers, which is permitted by my regex filter:
when regexp_instr(trim(atl.original_price),'[^0-9.$,]') .
I wrongly expected that a non numeric string like "$" would evaluate to NULL and then the to_number function would = NULL . But from experimenting it seems that it needs at least one number somewhere in the string. Otherwise it reduces the string argument to an empty string prior to running the to_number formatting and chokes.
For example select to_number(trim('$1'::text),'FM999999999999D00') will evaluate to 1 but select to_number(trim('$A'::text),'FM999999999999D00') will throw the empty string error.
My fix was to add an additional regex to my initial filter:
and regexp_instr(atl.original_price2,'[0-9]')>0 .
This ensures that at least one number will be in the string and after that the empty string error went away.
Hope my learning experience helps someone else.
I'm getting a error I can't quite explain I have a Excel list I want to load into memory, to see if the next row is still a relative row I check if the cell has a value by doing If value = "" Then but the value is 1012738 and it gives me a unhandled exception...
I can't quite understand why the code is giving a error, the cell value is formatted just like all previous cells that were checked. But here a error is thrown.
Maybe i'm just not seeing it, and someone can explain?
You should check the value each time.
Dim o As Object = oSheet.Range(xxx).Value
If (o IsNot Nothing) Then
Select Case o.GetType
Case GetType(Double)
' do work here
Case GetType(Integer)
' do work here
...
End Select
Else
...
End If
Your image isn't showing up for me.
Most exceptions in excel are datatype related. Most likely, you either have a NULL or a string that looks like an integer to the human eye. You can blindly try using trim(), int() or str() etc as needed to make sure you're actually testing an integer or matching a string if thats what you are about or you can test them in a programmatic method.
First, the programmatic method of testing ... isEmpty or isNull are needed to ensure the cell is good ...
if isEmpty(value) Then
<do something>
This will most likely catch the error which is causing the message. If empty is failing, try testing with isNull. One means the cell is empty, the other means that it wasn't initialize (rarely an issue in excel, but if using vba code it can happen).
Also ... your if statement is setup in a less than optimal... used <> in place of if then + else ...
Your code ...
If value = "" Then
General formula when you want to test that something is not something else, use the not equals ...
if value <> "value" Then
"some operation"
Column_A Column_B New (Expecting result for this situation)
#N/A #N/A Manual Posting
My function has problem for the last syntax"IFERROR(IFERROR(....)" . Currently, I get #N/A for the "New" column. However, I want to get "Manual Posting" instead.
My syntax:
=IF(OR(IFERROR(B1,A1)="Bank BPH",IFERROR(B1,A1)="GE Budapest Bank"),"GECapital",IF(IFERROR(B1,A1)="Avio Aero","GE Aviation",IFERROR(IFERROR(B1,A1),"Manual Posting")))
Ok, working it out I think I can explain.
IFERROR(x,y) returns value x, unless it's an error, then it returns y .. (even if it is an error).
You then take result of that, and compare it to a string:
IFERROR(A1,B1)="Bank BPH"
Assuming valid values, that expression will, of course, return TRUE or FALSE.
If both A1 and B1 are error, however, what happens ??
What is the result of:
#ERR="string" ??
answer: an error ...
so what does the IF do with an error? it's neither true, nor false.
You can simplify the situation to just this expression to see what's going on:
=IFERROR(A1,B1)="Bank BPH"
it returns an error.
Neither true nor false.
You're going to need another check condition for an error and how to handle it ..
perhaps:
=IF(AND(ISERROR(A1),ISERROR(B1)), "Manual post", IFERROR(A1,B1)="Bank BPH")
might do the trick ??
I think I figure out the correct syntax. I used to put the "AND(ISERROR(B18),ISERROR(A18)),"Manual Posting" to the end. However, I moved it at the beginning, it seems the problem can be solved.
=IF(AND(ISERROR(B18),ISERROR(A18)),"Manual Posting",IF(OR(IFERROR(B18,A18)="Bank BPH",IFERROR(B18,A18)="GE Budapest Bank"),"GE Capital",IF(IFERROR(B18,A18)="Avio Aero","GE Aviation",IFERROR(B18,A18))))
This seems simple, but evidently incorrect. Have any ideas?
The data:
Cell J5 value is simply this URL:
www.url.com/at/
The logic: Check if cell J5 ends in "/at/" or if it ends in "de/", value if true for either one of those is 1, value if false for both is zero.
Here's the function I'm trying out:
=IF(OR(RIGHT(J5,4)=“/at/"),(RIGHT(J5,3)=“de/"),"1","0")
My result is #NAME?
The double bracket characters are incorrect (before the /at and before the de/). Make sure you use "" around string literals. Secondly the closing bracket is not required after the first RIGHT and does not need open bracket before the second right. With these corrections the formula becomes:
=IF(OR(RIGHT(J5,4)="/at/",RIGHT(J5,3)="de/"),"1","0")
Your command has incorrect parenthesis.
Your command:
IF(OR(RIGHT(J5,4)=“/at/"),(RIGHT(J5,3)=“de/"),"1","0")
OR(RIGHT(J5,4)="/at/") is the cause of output #NAME because OR in this case has only one operand.
You should not close the parenthesis here and should close it after the second operand.
Try:
IF(OR(RIGHT(J5,4)=“/at/",RIGHT(J5,3)=“de/"),"1","0")
After entering any characters, the value which is stored in variable A is "0". Can anyone please assist me where I am going wrong as it is working fine if I enter number
Rookie
Public Sub MyFirstProgram()
Dim A As String
A = Val(InputBox("Enter your name", "NAME"))
MsgBox "My name is " & A
End Sub
The Val function converts a string to a Double numeric type.
Presumably, the names you are entering cannot be converted to a valid number, so the result is 0.
http://office.microsoft.com/en-us/excel-help/HV080557263.aspx
The Val function stops reading the string at the first character it can't recognize as part of a number.
So if you do something like =Val("123steve") it will return the numeric component: 123, but if you do =Val("Ebeneezer Scrooge") it stops, per the above remark -- since no characters have been converted to numeric value, it returns 0.