RE: [PHP-DB] Transaction suddenly not working

> --- "Ford, Mike" <M.Ford@leedsmet.ac.uk> wrote: > > > 0: INSERT INTO Table1 (LurkID, ProfileName, Edu, > > > WorkAuth, WorkExp, CarLev, Secu, Confi, Relo, > > > Telecomu, City1, State1, City2, State2, > > TravelPref, > > > SalaryAnnual, SalaryHourly, Available) VALUES (47, > > > 'ffffffff', 7, 2, 1015, 5, , '', '', '', > > > 'fds', 5, > > '', > > > > You have 2 successive commas in this last line, with > > no value in between -- > > this is your syntax error. > > > > > , 4, 26, , 2004-02-02)01062 : Duplicate entry > > '0-f' for key 1 > > > > And again on this line -- twice, in fact, since the > > previous line ends with > > a comma, too. > > > Mike are you saying that I should put single quotes > around all values regardless of the type ? > The double commas are happening because around an int > I don't use 'value' the single quotes . ? Don't know about other database engines, but in MySQL, it doesn't matter if an int is inserted as a string, i.e., with the quotes. My recollection is that MySQL handles the type conversion internally. So, assuming you are validating the entries before inserting, if a value is missing, assign to it ''. Whatever the field has as a default value will then be inserted into the table. dave