I am sure that this is not a good idea!
You are creating 7 forms with the same id-name! Suggest you move the
form creation outside the loop and make your submit button provide the
correct duty id.
[code]
<?php echo $form->create('Duty', array('action' =>
'displayDutyDetails')); ?>
<table cellpadding="0" cellspacing="0">
<?php foreach($duties as $duty): ?>
<tr>
<td><?php echo $duty['Area']['area']; ?></td>
<td><?php echo $duty['Station']['station_name']; ?></td>
<td><?php echo $duty['Bcu']['bcu']; ?></td>
<td><?php echo $duty['Duty']['duty_date']; ?></td>
<td><?php echo $duty['Duty']['on_duty']; ?></td>
<td><?php echo $duty['Duty']['off_duty']; ?></td>
<td><?php echo $duty['Duty']['remarks']; ?></td>
<td width="40px"><?php echo $form->submit('../img/icons/view.png',
array('label'=>'View', 'duty_id' => $duty['Duty']['duty_id'],
'width' => '40px', 'height' =>'25px',
'onmouseover' => 'javascript:this.src="../img/icons/
view_over.png"',
'onmouseout' => 'javascript:this.src="../img/icons/view.png"'));?
>
</td>
</tr>
<?php endforeach; ?>
</table>
<?php echo $form->end();?>
[/code]
Observe that the table is inside the form. You had a mixup with the
sequence of the table and the form tags.
Enjoy,
John
On Jan 14, 11:57 am, Bailey <baileys...@gmail.com> wrote:
> Hi all,
>
> I'm trying to display data in a html table from a database query.
>
> The data appears but I want to be able to provide an option from each
> record shown in the html table, i.e. (view more details button).
>
> I do this by creating a form inside the php loop that generates the
> table data. The form holds a hidden field that stores the record id i
> want to look up further with another function in the controller. In
> firefox this works perfectly but in IE it screws up, the button's dont
> appear and I get extra tables appearing above the results table that
> matches the same number of records in that table.
>
> (i.e. loop goes through 7 times, prints out 7 lots of data, but prints
> out 7 extra tables above).
>
> I'm sure theres a simple fix to this but i'm having real difficulty
> finding it. Does anyone know of anything i can do to fix this (or if
> theres another means that I can achieve the same result)? my table
> code is below, i'm not using any other styling at the moment apart
> from the cakephp default one.
>
> Many thanks!
>
> Code - View.ctp
> ==============================================================================
>
> <?php foreach($duties as $duty): ?>
> <?php
> echo $form->create('Duty', array('action' => 'displayDutyDetails'));
> echo $form->hidden('duty_id',array('value' => $duty['Duty']
> ['duty_id']));?>
>
> <table cellpadding="0" cellspacing="0">
> <tr>
> <td><?php echo $duty['Area']['area']; ?></td>
> <td><?php echo $duty['Station']['station_name']; ?></td>
> <td><?php echo $duty['Bcu']['bcu']; ?></td>
> <td><?php echo $duty['Duty']['duty_date']; ?></td>
> <td><?php echo $duty['Duty']['on_duty']; ?></td>
> <td><?php echo $duty['Duty']['off_duty']; ?></td>
> <td><?php echo $duty['Duty']['remarks']; ?></td>
>
> <td width="40px"><?php echo $form->submit('../img/icons/view.png',
> array('label'=>'View', 'width' => '40px', 'height'
> =>'25px',
> 'onmouseover' => 'javascript:this.src="../img/icons/
> view_over.png"',
> 'onmouseout' =>
> 'javascript:this.src="../img/icons/view.png"'));?>
> </td>
> <?php echo $form->end();?>
> </tr>
> <?php endforeach; ?>
> </table>
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