What I ment was that when using the GROUP by, you will see one of the
duplicates, and if you output the 'cnt' value you would see how many times
it has a duplicate. But I want to see all duplicate records in a table, so I
can add a delete button to the row.
If I have the email address as a PRIMARY i would very much like to know how
I would handle the input by the user when it is send to the database. I
assume it will generate an error upon submission. How do I go about letting
the user know they have entered an already existing address?
"Ignatius Reilly" <[EMAIL PROTECTED]> wrote in message
> I do not understand your first question.
> Your query in effect shows duplicates.
> To prevent future duplicates, you can do it in two ways:
> 1. with MySQL:
> create a unique index on what should be unique, most likely the email (it
> looks like your business logic implies one entry per email)
> in fact you would not have this problem if you had defined email as
> But you will have to remove the duplicates first - an unpleasant job.
> AFAIK, there is no easy way to do it with MySQL (or SQL in general). You
> have to do it procedurally by calling the entire table row by row.
> 2. Procedurally:
> validate user data before entry in the DB
> ----- Original Message -----
> From: "Marco Alting" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, December 04, 2002 11:47 AM
> Subject: [PHP-DB] displaying duplicate records...
> > I have a database which allows people to upload info and foto's. There's
> > unique ID field, but some people tend to upload their info more than
> > (its a contest site). What I'm able to do is to see how may duplicates
> > are using the following statement:
> > $query="SELECT COUNT(*) as cnt, voornaam,achternaam,leeftijd,ID,email
> > modellen GROUP BY email HAVING cnt >1";
> > But this only gives me numbers. Does anyone know how to display every
> > that has multiple duplicate ?
> > Or is there an elegant way to stop people from entering their info more
> > once?
> > --
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