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.
-Stuart
--
Stuart Dallas
3ft9 Ltd
http://3ft9.com/
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