I'd like to create a submit button that does NOT have a <div> around it. I assumed using the inputDefaults would make this happen like it does for all of the forms inputs, but - no luck.
Obviously I could just create a submit button via HTML, without CakePHP, but - I was hoping there'd be a cake answer. Here's what I've tried:
$this->Form->create(false, array('inputDefaults' => array('div'=>false)));
$this->Form->end('Submit');
echo $this->Form->submit('Submit', array('div'=>false));
Should do what you are after. The other example may have been Cake 1.2 or something; not sure.
It also appears you can just do this instead:
<?php
$options = array(
'label' => 'Update',
'value' => 'Update!',
'div' => false
)
);
echo $this->Form->end($options);
That looks more cakey.
Related
I'm developing a website using cakephp 2.x.
Now, I create a form using Cakedc/search. This form has input(select/dropdown list).
But the list is too long, so I want the dropdown to be view as unordered list (< ul >< li >).
Like in the lazada (search for brand): http://www.lazada.com.my/womens-watches-bags-accessories/.
the code:
<?php echo $this->Form->create('Product', array(
'url' => array_merge(array('action' => 'search'), $this->params['pass'])));
echo $this->Form->input('brand_id', array('label' => 'Brand', 'options' => $brands, 'empty' => 'Select Brand'));
<?php echo $this->Form->submit(__('Search', true), array('div' => false));
echo $this->Form->end();
?>
Please someone help me. Thanks in advance..
You'll have to use javascript to get the id of the clicked li element or the link inside, use data attributes for that for example. Then set this value to a hidden form field or directly submit the whole form using ajax to update your results.
Your example URL is a simple list by the way that is just doing redirect on click.
I have a form which has some disabled fields, when the form is submitted both $this->request->data and $_POST is empty, removing the disabled fields and it is fine again. I would have though it would still pass though the non-disabled fields. I've even tried to remove the disabled field attribute when the submit button is pushed but this still returns an empty array.
Is there something cake related that might be causing this?
Thanks
// SNIPPET FROM THE VIEW CODE:
$this->Form->create('Card', array('class' => 'GeneralValidate'));
$this->Form->input('Card.property_id', array('type'=>'select', 'empty'=>true , 'class' => 'required adminOnlyField', 'div' => array('class' => 'required')));
$this->Form->input('Card.building_id', array('type'=>'select', 'empty'=>true, 'id' => 'BuildingSelector', 'class' => 'adminOnlyField', 'label' => 'Building (If Applicable)'));
$this->Form->input('Prospect.waiting_list_details', array('value' => $prospect['Prospect']['waiting_list_details']));
$this->Form->input('SaleDetail.property_sold', array('class' => 'checkbox', 'checked' => $ps_checked));
$this->Form->input('SaleDetail.date_conditions_met', array('type'=>'text', 'class' => 'text date_picker adminOnlyField', 'value' => $this->Date->format($saledetail['SaleDetail']['date_conditions_met'])));
$this->Form->button('Save & Continue', array('type'=>'submit', 'label' => 'Save', 'name' => 'quicksave' , 'class' => 'submit long clear_ready_only'));
// JS FROM THE VIEW
$(function () {
var $adminOnly = $('.adminOnlyField');
$adminOnly.prop('disabled',true).prop('readonly',true);
$adminOnly.attr("onclick","return false");
$adminOnly.attr("onkeydown","return false");
$adminOnly.removeClass('required');
$adminOnly.removeClass('date_picker');
$('.clear_ready_only').click(function(e)
{
e.preventDefault();
$adminOnly.prop('disabled',false).prop('readonly',false);
$adminOnly.attr("onclick","return true");
$adminOnly.attr("onkeydown","return true");
$('#CardModifysaleForm').submit();
});
});
That's the way HTML works, disabled don't get posted. CakePHP can't change what is sent from the browser. If you still want the value you can set it as a hidden element.
Update
Some problems I see:
Missing Form::end() in view (always a good idea to insert it).
You never said your form was submitted from JS, first test with a simple form POST then JS.
Your JS code is set to submit a form by ID CardModifysaleForm. There's no such ID in your supplied view code and you're not setting your form to that ID from the snippet you supply.
I ended up removing the disabled option from this, leaving the ready only and added some addition CSS stylings so it looked disabled to the user. This is not the exact answer to the question but works as a different approach.
I've searched the web and have come up with nothing. (Multiple search engines too - I have looked!)
I'm trying to have a text link as the 'form submit' button. Any ideas if this is possible in CakePHP?
Current view code below!
<?php
echo $this->Form->create('trainees', array(
'action' => 'reassign'
));
echo $this->Form->input('emailaddress', array(
'value' => 'scott#something',
'type' => 'hidden',
));
echo $this->Form->submit('Re-Assign Mentor', array(
'class' => 'submit mid',
'before' => '<p>',
'after' => '</p>'
));
echo $this->Form->end();
?>
You need to use the HtmlHelper to output a link. In it's simplest form you use the text you want displayed with the URL that it should link to. In this case it will be JavaScript:
$this->Html->link('Submit Form', 'javascript:document.forms["myform"].submit();');
There are two additional parameters (a $options array and $confirmMessage boolean), but they along with the URL are optional.
You can also call your own JavaScript function if you need to do client side verification and call the submit function from there (also verify on the server as clients can lie).
http://book.cakephp.org/2.0/en/core-libraries/helpers/html.html#HtmlHelper::link
I am developing an application with CakePHP 2.3.2 and I am having some trouble with an input select on a form. I am creating an array, in my Controller, which contains a list of states. In my View I find that when I use this variable in the 'options' field of the input I do not get any select options. If I do a print_r on the variable, in the view, I see exactly what I think I should be seeing for the 'options' field. I have even tried copying this print_r output and putting it in the 'options' field and then the input select works fine.
Here is what I have
In Controller
$options = 'array(1 => \'NSW\',2 => \'ACT\',3 => \'NT\');
$this->set('all_states, $options);
In View
<?php
$options = $all_states;
echo $this->Form->create('Refine', array('url => '/ServiceDirectoryResults/view/refine'));
echo $this->Form->input('field' ,array(
'type' => 'select',
'label' => false,
'options' => $options
));
echo $this->Form->end('Refine Search');
?>
When I run this I see a select with no select options
If I add print_r($options) after the echo $this->Form->end('Refine Search'); I see
array(1 => 'NSW',2 => 'ACT,3 => 'NT')
Which is what I would expect as it is the content of the $options variable which was the $all_states variable passed from the controller. If I take this output from the print_r and replace the $option with it in the input select the select drop down works fine and I see the three options. For some reason I can't work out the select is working fine if I hard code the select options but it will not work if I pass a variable containing the array to the input select.
I would really appreciate if if someone could give me a clue what I am doing wrong here.
Kind Regards
Richard
you might try it like below:
echo $this->Form->input('field', array('type'=>'select','label' => false,
'options' => $options,'default'=>'2'));
to the following HTML being generated:
<option value="2" selected="selected">ACT</option>
option two is shown instead any other one.
Likely issue:
Arrays should not be made as strings like you have:
$options = 'array(1 => \'NSW\',2 => \'ACT\',3 => \'NT\');
Instead, just make an array:
$options = array(1 => 'NSW', 2 => 'ACT', 3 => 'NT');
Other notes:
Why are you setting $options to $all_states only to set it back?
Missing quotes all over - make sure if you start quotes, that you also end them
not good practice to hard-code your URLs (like in your Form->create)
I have a product search page with the form below. The search result is displayed on the same page with search bar at the top.
echo $this->Form->create('Searches', array('action'=>'products', 'type' => 'get', 'name' => 'textbox1'));
echo $form->input($varName1, array('label' => false));
echo $form->end('Locate');
I also have a little box next to the search result that allows (it doesn't work yet) the user to flag using checkboxes a product and accordingly update its database (table products and using model Product) with a button click. Note that I have a Searches controller for this search page.
<form method="link" action="/myapp/product/test_update_db>
<label><input type="checkbox" name="flag1" <?php echo $preCheckBox1; ?>>Flag 1</input></label>
<label><input type="checkbox" name="flag2" <?php echo $preCheckBox2; ?>>Flag 2</input></label>
<input type="submit" value="Update">
</form>
I'm having difficulty with this approach figuring out how to perform this check-box-and-DB-update routine. I'm getting to the link I'd like to go (/myapp/product/test_update_db), but I don't know how to take variables flag1 and flag2, along with row ID of this result ($results['Product']['id'])) to the new page.
Could someone guide me on how to perform this neatly? Is this general approach correct? If not, what route should I be taking? I'd prefer not to use javascript at this time, if possible.
EDIT: I think I can make this work if I use the URL for passing data.. but I'd still like to know how this could be done "under the hood" or in MVC. I feel like I'm hacking at the CakePHP platform.
UPDATE: So, I ended up using the URL parameters for retrieving information pieces like flag1 and flag2. I'm still looking for an alternative method.
To see where your is-checkbox-checked data is located, do the following in your controller:
// Cake 2.0+
debug($this->request->data);
// previous versions
debug($this->data);
If you want to pass data to your search controller from the current page, you can always add the data to your form:
$this->input
(
'Product.id',
array
(
'type' => 'hidden',
'value' => $yourProductId
)
);
I ended up using information embedded in the URL for getting submission data. Something like below..
In Products controller, when the form with flag1 and flag2 are submitted:
public function test_update_db() {
// Get variables from URL, if any, and save accordingly
$result = $this->Product->updateProduct($this->params['url'], 'url');
if ($result) {
$this->Session->setFlash('Successfully updated!', 'default', array('class' => 'success'));
$this->redirect($this->referer());
}
else {
$this->Session->setFlash('Update was unsuccessful!', 'default', array('class' => 'error'));
$this->redirect($this->referer());
}
}
This works for doing what I needed to do. I feel like there's a more proper way to do this though.
if ($result) {
$this->Session->setFlash('Successfully updated!', 'default', array('class' => 'success'));
$this->redirect($this->referer());
}