I have tried to pass $boxNumber as other numbers but the insert statment is still throwing an exception.
The single row/record in the table has an ID of zero, and remember ID auto increments. So the second record I try to insert throws an exception, that I'm trying to insert another row/record with a duplicate ID of zero.
Notice I am not including the ID in the insert statement, that should automatically be provided and incremented, I thought.
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by: Michael701Posted on 2009-07-04 at 07:43:25ID: 24777029
where did you set $boxNumber
try hard coding the boxNumber (just once) and see if you still get the error. This would tell you that's it's something in the insert function.