I use filtration form and pagination on same page to display great number of objects.
If i use POST form, after going through pagination filtration resets. If i use GET form filtration works, but URL is not clear and even have token.
Something like this:
?form[date_from][year]=&form[date_from][month]=&form[date_from][day]=&form[date_to][year]=&form[date_to][month]=&form[date_to][day]=&form[email]=email&form[submit]=&form[_token]=Nk0prilVJiROZaQQKvCt-hRfnKdh0IdDOWOIer
Is any way to make url more clear, hide token and unused parameters?
Well, i find no solution and write service for handling form filtration data in sessions.
Save filtration: use an array in controller:
$filter_array=[];
if ($search_form->isSubmitted() && $search_form->isValid()) {
$search_form->getData();
//save post/get form data without empty fields
$filter_array[$search_form->getName()]=array_filter(
$request->get($search_form->getName()),
function($value) { return $value !== ''; });
//... etc
}
Solution 1- make forms in your page links and add hidden fields with $filter_array
Solution 2- in page links use a script to add a hidden form field with page number and submit the form, simple but the form could be modified before click.
Solution 3- Put the paginator navigation and search results in a div, use jquery.load() for all actions (send form, page links, $( document ).ready() ...). The url will look like domain/page/#page_1:
{% for i in range(1, paginator.getTotalPages()) %}
<li {% if paginator.getPage() == i %} class="active"{%endif%}>
<a href="#page_{{i}}" onClick="
$('{{selector}}').load('{{ paginationPath }}',
{{filter_array|merge({page: i,})|json_encode()}});"
href="#page_{{i}}">{{ i }}</a>
</li>
{%endfor%}
Related
I need display particular fields in form in Symfony 2.6. I use a class Form. I have the folowing fields: name, email, message, send, recet. I need display all of them except recet .I try like this:
{{form_start(form)}}
{{ form_errors(form) }}
{{form_row(form.name)}}
{{form_row(form.email)}}
{{form_row(form.message)}}
{{form_end(form)}}
But, it's displaying all fields in form, it is not what I want. Even if i leave only {{form_start(form)}} and {{form_end(form)}} - its display all fields. Can someone help me wit this problem?
remove form_end, just close form with HTML
</form>
but then you must handle CSRF token generation by adding:
{{ form_widget(form._token) }}
{{ form_widget(form._token) }}
or try setting field you do not want to show using:
{% do form.recet.setRendered %}
but probably best way is not to add this field in the first place, rather than hiding it, by form options or event listeners depending on some criterias
A better solution is to remove the field from the form type. If you only remove it from the view, it may be interpreted as a blank submission for that field and delete existing data.
If you only use the form in one place, then just remove the field from the type. If you use it in multiple places, then you can selectively remove the field in a FormEvents::POST_SET_DATA event listener.
I'm using Laravel 4, and I have two tables related 'Many to many': 'Actividad' and 'Material'. Every Actividad can have one or more Materials, and every Material can belong to one or more Actividad.
So, I have made a form to create a new Actividad where you can save one or more Materials. I've done that with a multiselect input. Like that:
{{ Form::label('material_id', 'Material necesario:') }}
<p>{{ Form::select('material_id', $material_id, Input::old('material_id'), array('multiple')) }}</p>
I don't know if I'm doing correctly but, before saving anything, my first problem is that I'm only obtaining one result. I suppose I should get every option 'checked' in the form... The relevant part of my Controller is:
$material = Input::get('material_id');
return var_dump($material);
I should obtain a list of options selected, but the result of that in my browser is:
string(1) "1"
This is the first time I'm working with tables related in this way, and probably I'm doing something wrong (in the form, in my controller, in my Models,...)
Thank you very much for your help!!
Just change your code to this
{{ Form::select('material_id[]', $material_id, Input::old('material_id'), array('multiple')) }}
I hope this helps.
if you are using custom response handlers on the client side such in the case of submitting info with AJAX, all you need to do is to simple add "[]" to the name of the select control.
e.g.
<select name="material_id[]" multiple>
This is the same case as with regular PHP. The other methods are required if you want Laravel to handle the form elements for you when rendering a new response/view. ( page reload ). This reload doesn't happen with REST requests.
i'm trying to realize a form that show a list of exercices and under each one of them, the input texte for the logged in user to answer ,i don't need embeded forms because i will not change the exercice text i will just save the responses for each exercice of the given exam and user.
in my database i have three tables :exam that has many exercices and table exercice and table response that has the following fields (user, exercice, response, date)
please am stuck help me !
You can insert a form when displaying your exercices unter each of them and create a route & controller for this form but the router/controller will receive as paramater the Id or Code of your exercice so that when it save the response it set directly it's exercice.
So you must get a ManyToOne relation between Exercice & Reponse.
Just tell me if i've not understood (my english is very poor )
I thik you must add dynamicly your form into the template not send via controller but the action of each form will be a controller route that will yse your POST data to edit exercice
route_name:
pattern: /patten/{id} #hier id is the id of the exercice
defaults: { _controller:.....:Response }
ResponseAction(Exercice $e}
{
In this controller you get the POst datas and you can add your response on the
article
}
Twig:
{{ exercice.content }} {# an example #}
{# a form that you will normaly generate will jQuery or another of your choice #}
<form action="{{ path('route_name',{'id':exercice.id}) }}" method="POST">
//
your fields ...
</form>
I have a simple search form I want to reuse across multiple pages in my codeigniter application. For example, right now I have a search form in the sidebar and I'm planning on displaying that sidebar on the index, about, and other pages.
I want to have the form validation display errors on the same page the users submits the form from.
For example:
User is on About page.
User submits form with invalid data
User sees error in the sidebar on the About page
and
User is on Index page.
User submits form with invalid data
User sees error in the sidebar on the Index page
But I'd like to reuse that form validation logic. I just want it to display the error on whichever page the user posted from.
Any ideas how to do that? Sorry for the noob question, I'm pretty new to CI.
Here you have to think globally.
Step.1 : Make one view file : display.php
which contains :
<div id = "main">
<div id = "header">
[load header file here]
</div>
<?php
if(validation_errors() != '') {
?>
<div id = "error">
<?=validation_errors()?>
</div>
<?php
}
?>
<div id = "content">
<?=$page?>
</div>
<div id = "footer">
[load footer file here]
</div>
</div>
Step.2 : About us Page.(controlller)
.... Your data ....
at end of controller function
$data['page'] = 'aboutus';
$this->load->view('display',$data);
With regards to your comment on the above question you could use Flash data
With the assumption that you have the session library loaded, here is a helper function.
function last_page($page = null){
$ci = get_instance();
if($page === null)
return $ci->session->flashdata('last_page');
$ci->session->set_flashdata('last_page', $page);
}
then you can call last_page('about'); at the top of the about page, and then when you want to find out what the last page you were on was you can just call last_page(); with no params.
In the User Guide, there's different ways to configure sets/groups of rules. Then you can simply have something like:
if ($this->form_validation->run('signup') == FALSE)
{
$this->load->view('myform');
}
else
{
$this->load->view('formsuccess');
}
Which will run your "signup" group of validations. To me, this is the cleanest way to achieve reusable validation rules.
This is a perfectly valid question.
I'm not a PHP expert, nor a CI expert, but the fact is sometimes you want to post to a controller that didn't create the view from which you're posting. Which means posting back to itself is not going to work.
I came across this post on the Ellislab forum:
http://ellislab.com/forums/viewthread/217176/
On this page, There are 2 methods of going about it. Both of which use flashdata and both of which are totally valid, IMHO.
The first: create a helper function
http://ellislab.com/forums/viewreply/1003010/
The second: extend the CI_Form_Validation Class.
http://ellislab.com/forums/viewreply/1047536/
The second is the way I went as it seems cleanest although some may argue whether the form validation class should know anything about flash data.
I'm using jquery-ui tab example to add extra tabs. I changed that code to be able to add tabs that load a form via Ajax. I was able to create that just changing these:
var $tabs = $( "#tabs").tabs({
cache: true,
tabTemplate: "<li><a href='formularioAgricola.php' id='#{label}'>#{label}</a> <span class='ui-icon ui-icon-close'>Remove Tab</span></li>"
//ajaxOptions: a
});
So I changed the tabTemplate to load the same Form always.
My problem is that I'm not sure how to retrieve, either to tell that every tag from that form use jquery-ui stuff, like buttons, datepickers, etc.
In a regular form I would do something like:
$("#btnRevisar").button()
But when we talk about form load via Ajax it is different.
and also, how can I try to differ one form from other one, if they are all named with the same name, is it possible?
Thanks guys
Carlos.
Within the tabs docs page, tab titled "Events" there is a "load" event. The "ui" argument gives you access to an object that includes the current panel that is loaded. If you are using same ID on forms, beware that ID's must be unique in a page.
var $tabs = $( "#tabs").tabs({
cache: true,
tabTemplate: "<li><a href='formularioAgricola.php' id='#{label}'>#{label}</a> <span class='ui-icon ui-icon-close'>Remove Tab</span></li>",
/* add new option for load event*/
load: function( event, ui){
var $currTabContentPanel=$(ui.panel);
/* only look in currently loaded content for formClass*/
$currTabContentPanel.find('.formClass').doSomething()
}
});