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 Dallas
3ft9 Ltd
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