2013/7/22 Stuart Dallas <stuart@xxxxxxxx>
>
> On 22 Jul 2013, at 08:04, Tamara Temple <tamouse.lists@xxxxxxxxx> wrote:
>
> > On Jul 22, 2013, at 1:19 AM, Karl-Arne Gjersøyen <karlarneg@xxxxxxxxx>
> 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
