Re: If( Query)

[Chris Hale <php@xxxxxxxxxxxxxxx>]

Niel Archer wrote:
> Hi
>
>   
> > I have the following function:
> >
> > function 
> > add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)
> >     {
> >     connect();
> >     if($item_pix == "")
> >     {
> >         $sql = "INSERT INTO items 
> > (item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES 
> > ('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id')";
> >     }
> >     else {
> >         $sql = "INSERT INTO items 
> > (item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES 
> > ('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id')";
> >     }
> >     mysql_query($sql);
> >     return;
> > }
> >
> > I am using the if statement because i want it so that if no picture is 
> > uploaded the entry is blank and the mysql database has a default entry 
> > of na.gif which is a "picture coming soon picture".
> >
> > It works fine when i run in localy on MAMP, but if i run it on my web 
> > server it doesnt add the row.
> >     
>
> You should be checking the mysql_query call for success and output the
> error if it fails.  Something like:
>
>    mysql_query($sql) or die('Insert failed: ' . mysql_error());
>
> You'll now why it's failing then.  Make sure you have error reporting
> enabled.
>
>   
> > Is this a compatability error? or is there a better way to write this?
> >
> >     
>
>
> --
> Niel Archer
>
>
>
>   
I have fixed it now:

function 
add_item($item_name,$item_desc,$item_price,$item_man_id,$item_cat_id,$item_pix)
   {
   connect();
   if($item_pix == "")
   {
       $sql = "INSERT INTO items 
(item_name,item_desc,item_price,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_man_id','$item_cat_id')";
   }
   else {
       $sql = "INSERT INTO items 
(item_name,item_desc,item_price,item_pix,item_man_id,item_cat_id) VALUES 
('$item_name','$item_desc','$item_price','$item_pix','$item_man_id','$item_cat_id')";
   }
   mysql_query($sql);
   return;
}

Thanks anyway.

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