L
Les
Newfld:Left([yourfield], InStr([yourfield], ".") -1)
everything to the left of the decimal point.
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K
Kirk P.
Your function works perfectly for those product numbers that have decimals in the string, however it looks like I have some product numbers that DO NOT have decimals. Such as
Produc
5461C00047.1IB
777777
In the case of 77777777, I get #Error returned as a value in the query. Any additional advice
----- Les wrote: ----
Newfld:Left([yourfield], InStr([yourfield], ".") -1
Produc
5461C00047.1IB
777777
In the case of 77777777, I get #Error returned as a value in the query. Any additional advice
----- Les wrote: ----
Newfld:Left([yourfield], InStr([yourfield], ".") -1
J
John Spencer (MVP)
You can force a decimal point at the end by concatenating it to the original
string. This may still give you errors if your field is a null.
NewFld: Left([yourfield], InStr([yourfield] & ".", ".") -1)
Or you can use an IIF statement
NewFld: IIF(Instr([yourField] & "",".")>1, Left([yourfield], InStr([yourfield] &
".", ".") -1),[YourField])
string. This may still give you errors if your field is a null.
NewFld: Left([yourfield], InStr([yourfield] & ".", ".") -1)
Or you can use an IIF statement
NewFld: IIF(Instr([yourField] & "",".")>1, Left([yourfield], InStr([yourfield] &
".", ".") -1),[YourField])
Your function works perfectly for those product numbers that have decimals in the string, however it looks like I have some product numbers that DO NOT have decimals. Such as:
Product
5461C00047.1IBA
7777777
In the case of 77777777, I get #Error returned as a value in the query. Any additional advice?
----- Les wrote: -----
Newfld:Left([yourfield], InStr([yourfield], ".") -1)
-----Original Message-----
I have part numbers like this 11A800.7632 or AMG7002488.A544
.
K
Kirk P.
Perfect - thanks for the help!!!
----- John Spencer (MVP) wrote: ----
You can force a decimal point at the end by concatenating it to the origina
string. This may still give you errors if your field is a null
NewFld: Left([yourfield], InStr([yourfield] & ".", ".") -1
Or you can use an IIF statemen
NewFld: IIF(Instr([yourField] & "",".")>1, Left([yourfield], InStr([yourfield]
".", ".") -1),[YourField]
Kirk P. wrote
----- John Spencer (MVP) wrote: ----
You can force a decimal point at the end by concatenating it to the origina
string. This may still give you errors if your field is a null
NewFld: Left([yourfield], InStr([yourfield] & ".", ".") -1
Or you can use an IIF statemen
NewFld: IIF(Instr([yourField] & "",".")>1, Left([yourfield], InStr([yourfield]
".", ".") -1),[YourField]
Kirk P. wrote
Your function works perfectly for those product numbers that have decimals in the string, however it looks like I have some product numbers that DO NOT have decimals. Such as
Produc 5461C00047.1IB
777777
In the case of 77777777, I get #Error returned as a value in the query. Any additional adviceNewfld:Left([yourfield], InStr([yourfield], ".") -1----- Les wrote: ----I have part numbers like this 11A800.7632 o AMG7002488.A54-----Original Message----I am looking for an Access function that would KEE everything to the left of the decimal point
Any ideas