2013/7/22 Stuart Dallas <stu...@3ft9.com> > > On 22 Jul 2013, at 08:04, Tamara Temple <tamouse.li...@gmail.com> wrote: > > > On Jul 22, 2013, at 1:19 AM, Karl-Arne Gjersøyen <karlar...@gmail.com> > wrote: > > > >> Hello again. > >> I have this this source code that not work as I want... > >> > >> THe PHP/HTHML form fields is generated by a while loop and looks like > this: > >> > >> <input type="number" name="number_of_items[]" size="6" value="<?hp > >> echo "$item"; ?>" required="required"> > >> > >> > >> the php source code look like this: > >> <?php > >> if(!empty($_POST['number_of_itemsi'])){ > >> include('../../connect.php'); > >> > >> foreach($number_of_items as $itemi){ > >> echo "$itemi<br>"; > >> > >> $sql = "UPDATE item_table SET number_item = '$item' WHERE date > >> = '$todays_date' AND sign = '$username'"; > >> mysql_query($sql,$connect) or die(mysql_error()); > >> } > >> } > >> > >> ?> > >> > >> The problem is: > >> Foreach list every items as expected in PHP doc and I thought that $sql > and > >> mysql_query should be run five times when I have five items. > >> But the problem is that only the very last number_of_items is written > when > >> update the form.. > >> I believe this is becayse number_of_items = '$item in $sqk override > every > >> earlier result. > >> > >> So my querstion is. How to to update the database in this case? > >> > >> Thanks again for your good advice and time to help me. > >> > >> Karl' > > > > Either the code you posted isn't the actual code, or if it is, the > errors should be rather obvious. Post *actual* code. > > The error is rather obvious: it loops around an array running an update > statement that will modify a single row in the table, so it's not > surprising that it appears like only the last entry in the array has been > stored. > > Yes, i know that only one a singe row is updated and that is the problem. What can I do to update several rows at the same time? Thank you very much for all your good adivce.
Karl