plague --
unfortunately i don't have the beginnings of this thread, but if your code
below is verbatim, it looks as though you are missing the mysql_query()
statement. that should drop your information.
kate
> -----Original Message-----
> From: plague [mailto:[EMAIL PROTECTED]]
> Sent: Friday, August 24, 2001 12:57 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Variables in MySQL Insert Queries
>
>
> I am using this code to connect to my database and insert form data from
> a
> user:
>
> $connect = mysql_connect("myhost","user","pass") or die (" not
> connected");
> @mysql_select_db("dbname");
> $sql="INSERT INTO tablename
> (id,first,last,age,email,sfuser,sfship,icq,ac,loca,ref)
> Values(,`$first`,`$last`,'$age',`$email`,`$sfuser`,`$sfship`,`$icq`,`$ac`,`$
> loca`,`$ref`)";
> echo (mysql_affected_rows()?"success":"failure");
> mysql_close($connect);
>
> The script returns "success" except for it doesn't insert the data (
> from a
> form ).
>
> The age column is BLOB, not INT.
>
> I would really appreciate the help of someone, as this has really
> stumped
> me.
>
> Thanks,
>
> plague
>
>
>
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
