I've got a complicated re-population system that works, but it's cumbersome.
I've got a dropdown menu for a country of birth, and when the users submit the form it gets thrown in a database. The dropdown menu then gets repopulated with whatever is in the database if there is something.
I get the value from the database like this:
$birthplacecountry = (!set_select('birthplacecountry') && $birth_citizenship) ? $birth_citizenship[0]['birthplacecountry'] : (set_select('birthplacecountry') ? set_select('birthplacecountry') : '');
That works fine, as I've then got my options looking like this:
<option value="Afganistan" <?php if ($birthplacecountry == 'Afganistan') echo 'selected="selected"';?>>Afghanistan</option>
<option value="Albania" <?php if ($birthplacecountry == 'Albania') echo 'selected="selected"';?>>Albania</option>
<option value="Algeria" <?php if ($birthplacecountry == 'Algeria') echo 'selected="selected"';?>>Algeria</option>
<option value="American Samoa" <?php if ($birthplacecountry == 'American Samoa') echo 'selected="selected"';?>>American Samoa</option>
However, it's annoying having to compare $birthplacecountry to the name of the value manually. Is there a general way to compare something to the options value?
Some sort of functionality that works like: if ($birthplacecountry == this.option.value)
In codeIgniter, you should use Form helper to produce a select/dropdown, i.e.
$options = array(
'Afganistan' => 'Afganistan',
'Albania' => 'Albania',
);
echo form_dropdown('birthplacecountry', $options, 'Albania');
In this example, a dropdown with name birthplacecountry will be created and selected option will be Albania, the seconf one. The third parameter sets the selected option, so, when you want to repopulate the form with selected option, you can do something like this
echo form_dropdown('birthplacecountry', $options, set_select('birthplacecountry', $this->input->post('birthplacecountry') );
Related
In Yii2 I'm trying to construct hidden input
echo $form->field($model, 'hidden1')->hiddenInput()->label(false);
But I also need it to have some value option, how can I do that ?
Use the following:
echo $form->field($model, 'hidden1')->hiddenInput(['value'=> $value])->label(false);
Changing the value here doesn't make sense, because it's active field. It means value will be synchronized with the model value.
Just change the value of $model->hidden1 to change it. Or it will be changed after receiving data from user after submitting form.
With using non-active hidden input it will be like that:
use yii\helpers\Html;
...
echo Html::hiddenInput('name', $value);
But the latter is more suitable for using outside of model.
simple you can write:
<?= $form->field($model, 'hidden1')->hiddenInput(['value'=>'abc value'])->label(false); ?>
You can do it with the options
echo $form->field($model, 'hidden1',
['options' => ['value'=> 'your value'] ])->hiddenInput()->label(false);
you can also do this
$model->hidden1 = 'your value';// better put it on controller
$form->field($model, 'hidden1')->hiddenInput()->label(false);
this is a better option if you set value on controller
$model = new SomeModelName();
if ($model->load(Yii::$app->request->post()) && $model->save()) {
return $this->redirect(['view', 'id' => $model->group_id]);
} else {
$model->hidden1 = 'your value';
return $this->render('create', [
'model' => $model,
]);
}
Like This:
<?= $form->field($model, 'hidden')->hiddenInput(['class' => 'form-control', 'maxlength' => true,])->label(false) ?>
You can use this code line in view(form)
<?= $form->field($model, 'hidden1')->hiddenInput(['value'=>'your_value'])->label(false) ?>
Please refere this as example
If your need to pass currant date and time as hidden input :
Model attribute is 'created_on' and its value is retrieve from date('Y-m-d H:i:s') ,
just like:"2020-03-10 09:00:00"
<?= $form->field($model, 'created_on')->hiddenInput(['value'=>date('Y-m-d H:i:s')])->label(false) ?>
<?= $form->field($model, 'hidden_Input')->hiddenInput(['id'=>'hidden_Input','class'=>'form-control','value'=>$token_name])->label(false)?>
or
<input type="hidden" name="test" value="1" />
Use This.
You see, the main question while using hidden input is what kind of data you want to pass?
I will assume that you are trying to pass the user ID.
Which is not a really good idea to pass it here because field() method will generate input
and the value will be shown to user as we can't hide html from the users browser. This if you really care about security of your website.
please check this link, and you will see that it's impossible to hide value attribute from users to see.
so what to do then?
See, this is the core of OOP in PHP.
and I quote from Matt Zandstr in his great book PHP Objects, Patterns, and Practice fifth edition
I am still stuck with a great deal of unwanted flexibility, though. I rely on the client coder to change a ShopProduct object’s properties from their default values. This is problematic in two ways. First, it takes five lines to properly initialize a ShopProduct object, and no coder will thank you for that. Second, I have no way of ensuring that any of the properties are set when a ShopProduct object is initialized. What I need is a method that is called automatically when an object is instantiated from a class.
Please check this example of using __construct() method which is mentioned in his book too.
class ShopProduct {
public $title;
public $producerMainName;
public $producerFirstName;
public $price = 0;
public function __construct($title,$firstName,$mainName,$price) {
$this->title = $title;
$this->producerFirstName = $firstName;
$this->producerMainName = $mainName;
$this->price = $price;
}
}
And you can simply do this magic.
$product1 = new ShopProduct("My Antonia","Willa","Cather",5.99 );
print "author: {$product1->getProducer()}\n";
This produces the following:
author: Willa Cather
In your case it will be something semilar to this, every time you create an object just pass the user ID to the user_id property, and save yourself a lot of coding.
Class Car {
private $user_id;
//.. your properties
public function __construct($title,$firstName,$mainName,$price){
$this->user_id = \Yii::$app->user->id;
//..Your magic
}
}
I know it is old post but sometimes HTML is ok :
<input id="model-field" name="Model[field]" type="hidden" value="<?= $model->field ?>">
Please take care
id : lower caps with a - and not a _
name : 1st letter in caps
I have have a view in which there's a form that manages products (either add new product or -if an id passed- editing an existing one). If an id is passed then the form action should be eg 'admin/product/manage/5', if no id passed then it should be like this 'admin/product/manage'.
<?php echo form_open('admin/product/manage/{optional product id}', array('class' => 'ajax-form')); ?>
I have also created and this route:
$route['admin/product/manage'] = "admin/product/manage";
$route['admin/product/manage/(:num)'] = "admin/product/manage/$1";
How can I make my form action work correctly? is it possible to put inside the action the route somehow??
This is my Controller:
public function manage($id = NULL){
//fetch a single product to edit or create a new one
if (isset($id) === true) {
$data['prod'] = $this->product_model->get($id);
$data['vers'] = $this->product_version_model->get_by('product_id',$id);
} else {
$data['prod'] = $this->product_model->make_new();// this returns $product->product_name = ''; in order to be empty the input field and not throughing errors
}
$this->product_model->save_product();
$this->product_version_model->save_version();
// load the view
$this->layout->view('admin/products/manage', $data);
}
This is my view:
<?php echo form_open('admin/product/manage', array('class' => 'ajax-form')); ?>
<p>
<label for="product_name">Product *</label>
<input type="text" name="product_name" value="<?php echo set_value('product_name', $prod->product_name); ?>" />
<?php echo form_error('product_name'); ?>
</p>
<?php echo form_close() . PHP_EOL; ?>
You need to declare both possible routes in order of importance, so:
$route['admin/product'] = "admin/product/manage";
$route['admin/product/(:num)'] = "admin/product/manage/$1";
From the Codeigniter Docs:
Routes will run in the order they are defined. Higher routes will always take precedence over lower ones.
Edit:
According to the changes you have made to your question I can say the following:
First of all isset() returns boolean only, so you don't need the type check "=== true". isset($id) is sufficient.
In order to have your form action set to the id you need to include it either in a hidden field or in the action itself.
So for example:
$action_id = (isset($id) ? '/'.$id : ''); // Using ternary operators here
echo form_open('admin/product/manage'.$action_id, array('class' => 'ajax-form'));
and add the id to the view data in your controller:
$data['id'] = $id;
As a side note: In order to comply with SoC (Separation of Concerns) you'd prepare all data in your controller (with e.g. models all having their own task) and pass the processed data to the view instead of partially generating data in the view itself.
I'm using cakephp 2.3.0. I searched in the manual for quite awhile, but I haven't found the answer. Also, I've searched the Internet, but still haven't found what I'm looking for. SO, I'm posting my question here. Note, I'm fairly new to cakephp.
Scenario:
I have a simple form with two fields: activity and zip code.
I'm using POST on the form.
When I type in some value in those fields and submit, I echo those 'post' values/parameters and display in the browser screen. What I typed in, I can see on the screen, but the number '1' is added to the end of what I typed in the form.
Here is an example. I type in these values in the form, 'walk' and '44555'. Then I click 'Submit'. The post goes to my controller's action, which then calls my view. My view is displayed on the browser screen and I echo out those 'post' values. The results on screen are 'walk1' and '445551'.
Example #2: If I follow the steps above and don't enter any values in my form (I'll add error checking later), what I see on the browser screen is '1' and '1'.
I am unable to figure out why I am getting the value of '1' added to my form's POST values?
I'll be glad to include any other additional php code to this posting, if requested by someone trying to help.
Here is my FORM code (from my view)...I know there are DIV helpers, but I'll get to that later:
echo $this->Form->create(null, array('url' => array('controller'=>'activities', 'action'=>'results'))); ?>
<div class="box1" style="position:relative; top:10px; left:10px; float: left;">
Search here.... <br>
<hr>
<?php echo $this->Form->input('activityName', array('size'=>'30',
'label'=>'Activity Name:', 'value'=>'i.e. walking, etc.'));?>
<br>
<?php echo $this->Form->input('zip', array('size'=>'7', 'label'=>'Postal Code:')); ?>
<br>
</div>
<div class="box1" align="right">
<?php echo $this->Form->end('Go Search');?>
</div>
Here is my controller code:
<?php
class ActivitiesController extends AppController {
public $helpers = array('Html', 'Form');
public function index() {
//other code....
}
public function results() {
$this->layout = 'second';
$name = $this->request->data['Activity']['activityName'];
$pCode = $this->request->data['Activity']['zip'];
$this->set('theName', $name);
$this->set('theZip', $pCode);
$this->set('results', $this->Activity->
find('all', array('conditions' => array('name' => $name, 'postal_code' => $pCode))));
$this->set('title_for_layout', 'Results');
$this->render();
}
}
?>
My final view code. I left off some of the code...just showing the part that matters:
<div style="position:relative; top:10px; left:5px; ">
<?php echo print_r($theName); ?>
<br>
<?php echo print_r($theZip); ?>
Thanks
The 1 comes from printing the return value of print_r() which is true (i.e. 1).
In other words: you shouldn't do echo print_r(), just do print_r(). The function handles the printing by itself, you don't have to print the results manually.
(Also, print_r() is almost never the best choice to print out values except when debugging and even then CakePHP's debug() is much more suitable.)
I have this view:
<?php echo form_open(); ?>
<?php echo form_input('username', '', ''); ?>
<?php echo form_submit('submit', 'Submit'); ?>
<?php echo form_close(); ?>
<?php echo validation_errors(); ?>
and this controller method:
function test() {
$username = $this->input->post('username');
$this->form_validation->set_rules($username, 'Username', 'required|min_length[4]');
if ($this->form_validation->run() !== FALSE) {
echo 'Tada!';
}
$this->load->view('test');
}
but when I leave the username field blank, nothing happens. However, if I type in something in it will tell me the field is required. I've downloaded and given CI a fresh install almost ten times now, trying to load with and without different helpers etc. It's becoming really frustrating, please help.
Try using this:
$this->form_validation->set_rules('username', 'Username', 'required|min_length[4]');
The problem lies in the first parameter of set_rules, which is the field name.
The $username you're passing is basically setting the field name to validate as whatever the user puts in the input field. If you were to type 'username' into the input box, you'd see that your form validates.
Change the line
$this->form_validation->set_rules($username, 'Username', 'required|min_length[4]');
to
$this->form_validation->set_rules('username', 'Username', 'required|min_length[4]');
Maybe is the problem of this line
<?php echo form_open(); ?>
If you leave it blank it basically send back to the controller itself and calling the construct and index function only. In this case your function dealing with form processing is "test()"
try this
<?php echo form_open('yourControllerName/test'); ?> //test is the function dealing with
if it is not working try on this
<?php echo form_open('test'); ?>
Im trying to render a row field in a template with some extra styles, like this:
<?php echo $form['email']->renderRow(array('class' => 'text')) ?>
<?php echo $form['email']->renderError() ?>
The problem occurs when my form doesnt validate on this field... then it displays the error message 2 times!, i.e the renderRow renders one errorMsg string, and the renderError does it again... How can i stop renderRow from displaying the error message?
If I just do this, then it works:
<?php echo $form['email'] ?>
But in that case I cant style the field as I want....
thanks!
I am pretty sure this is also valid for 1.2. Instead of using renderRow, use something like this:
<?php echo $form['FormElementName']->renderLabel() ?> //display form element label
<?php echo $form['FormElementName']->renderError() ?> //display form element error (if exist)
<?php echo $form['FormElementName']->render(array('class' => 'text')); ?> //display form element
renderRow does them all at once.
EDIT From comments (Flask) - added ->render(array('class' => 'text'));