D damcloodt Registered User. Local time Today, 14:56 Joined Sep 23, 2008 Messages 16 Apr 3, 2009 #1 Hello, I want to delete the third digit for all records in a column, is there a query that I can use in design view to do this? for instance "ab 123" should become "ab123" Thanks a lot for all your help!
Hello, I want to delete the third digit for all records in a column, is there a query that I can use in design view to do this? for instance "ab 123" should become "ab123" Thanks a lot for all your help!
DCrake Remembered Local time Today, 13:56 Joined Jun 8, 2005 Messages 8,620 Apr 3, 2009 #2 if you actually need to delete it see other post use the following in an update query Code: UPDATE YourTableName SET [YourFieldName] = Replace([YourFieldName]," ","");
if you actually need to delete it see other post use the following in an update query Code: UPDATE YourTableName SET [YourFieldName] = Replace([YourFieldName]," ","");
gemma-the-husky Super Moderator Staff member Local time Today, 13:56 Joined Sep 12, 2006 Messages 16,075 Apr 3, 2009 #3 dc's method will delete any spaces anywhere if you want to specifically delete the third character then you need to select the first two chars, and the 4th char to the end so mystring = left(mystring,2) & mid(mystring,4)
dc's method will delete any spaces anywhere if you want to specifically delete the third character then you need to select the first two chars, and the 4th char to the end so mystring = left(mystring,2) & mid(mystring,4)
D damcloodt Registered User. Local time Today, 14:56 Joined Sep 23, 2008 Messages 16 Apr 3, 2009 #4 gemma-the-husky said: dc's method will delete any spaces anywhere if you want to specifically delete the third character then you need to select the first two chars, and the 4th char to the end so mystring = left(mystring,2) & mid(mystring,4) Click to expand... thanks it worked!!
gemma-the-husky said: dc's method will delete any spaces anywhere if you want to specifically delete the third character then you need to select the first two chars, and the 4th char to the end so mystring = left(mystring,2) & mid(mystring,4) Click to expand... thanks it worked!!