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I have db (mySql ) with 3 records.
When i run query to make list of all those records ( where id = 10 ) i get one record corectly displayed but instead second record i get error message:
Warning: mysql_fetch_assoc(): 4
Why?
p.s. Everything else is working fine ( connections, selects and so on )..
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I am connected to db and i can work with db....
2.) Table creation:
Maybe dump question... Since i have 3 records (2 of them should mach the query ), is it possible that there is a problem with "dinamic table creation" , becuse i allways get one record displayed(regardless how many records are in DB)...
you are free-ing up the recordset aftering fetching the first row of the data... then you are trying to fetch the remaining rows in the do-while loop.
IMHO, you shold improve your coding style... try to study few codes posted on the forum or take some reference from a book.