I'm having a little problem regarding a CI Form:
I retrieve some entries from the db and put them in hidden fields in my form. One of the values isn't filled into the hidden field, even though I can verify it's there:
<?php
echo $grno;
echo $dsca;
$hidden = array(
'grno' => $grno,
'dsca' => $dsca,
'bsex' => 'männlich',
'noch' => '0',
....
the echo statement shows both values in the view but the hidden field that is supposed to be filled with the value of $grno is empty..
Any ideas?
Thanks a lot, this is driving me crazy...
Regards
Lutz
Related
This page actually a preview where user can't change anything,that he has given before.I have tried bellow code,
echo $this->Form->input('exchange_type', array(
'disabled' => 'disabled',
'empty' => '--Please Select--',
'options' => array(
'6' => 'POINT_TO_PRODUCT',
'7' => 'POINT_TO_GIFT',
'2' => 'POINT_TO_GAME'
)
));
Here field has disabled but it's sending null value to database.I am trying to send actual value that user has been selected.How can I do this ?
That's how HTML works, values of disabled elements are not being sent.
What you can do is using a hidden field, that's what the form helper automatically does when using for example checkboxes, in order to ensure that there's always a value being sent, as unchecked checkboxes do not submit any value, just like disabled inputs.
The hidden field should have the same name as the actual field, and it should be placed before the actual field, that way the hidden value will only be sent in case the following element is disabled.
echo $this->Form->hidden('exchange_type');
echo $this->Form->input('exchange_type', array(
'disabled' => true,
// ...
));
That would pick up the previously POSTed value for both the hidden input and the select input, and the hidden input would be submittable.
See also Cookbook > Helpers > FormHelper > FormHelper::hidden()
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 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'm using this way to set the birthday in CakePHP, but when i turn back in the page I always get the current day date, so if I save the page without touch it, i get the today date as birthday:
<?php
$attributes = array (
'minYear' => date('Y') - 100,
'maxYear' => date('Y') - 0,
'label'=> false,
'default' => $user['Profile']['birthday'],
'value'=>$user['Profile']['birthday'] // how to set the previous saved data?
);
$options = array (
'id' => 'birthday',
'after' => '<div class="message">Inserisce la tua data di nascita</div>'
);
echo $this->Form->input('Profile.birthday', $attributes, $options);
?>
How can i set the previous saved data from the database to the form?
From your question, I assume that you are writing some kind of Edit-Form. In this case, CakePHP should automagically set the saved form data in your inputs. You should be able to savely discard default and value in your $options-array.
Debug $this->form->data to see whether the data is there. Otherwise you should check your controller code.
By the way: the signature of the input function is public function input($fieldName, $options = array()), so you probably want to merge the values in your $options-array to your $attributes-array.
This type of input stores data in this format:
your can put the value in controller this format:
$this->request->data['Profile']['birthday']['year'] ='xxxx';
$this->request->data['Profile']['birthday']['month']='xx';
$this->request->data['Profile']['birthday']['day'] = 'xx';
It will automatically display as selected in view.
Ask if not work for you
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());
}