Loading a codeigniter view with a form from inside another view - codeigniter

Using codeigniter, I've been trying to load a view inside of a foreach loop, as follows:
$posts = $this->postslibrary->getAllPosts();
foreach($posts as $post){
$home['content'][$i] = $this->load->view('post', $post['data'], true);
$i++;
}
$this->load->view('head');
$this->load->view('home', $home);
$this->load->view('footer');
Each of those post views looks a little like this:
<div class="postnum<?=$post_num?>">
<p>Posted by: <?=$poster_name?></p>
<p>Reply to: <?=$poster_name?></p>
<form>
<input type='text' />
<input type='submit' />
</form>
</div>
And they're being loaded mostly successfully in the 'home' view (which is below for thoroughness).
<div ="posts">
<?php
for($i=0;$i<$count;$i++)
{
echo($content[$i]);
}
?>
<div class="clear"></div>
<a href='/posts/browse/'>Load more items</a>
</div>
But my output ends up looking like:
<div class='posts'>
<div class='postnum1'>
<p>Posted By: Jim</p>
<p>Reply to Jim</p>
<input type='text' />
<input type='submit' />
</div>
</div>
Why are my form tags not coming through?

Check if you already have a form around the current form. Chrome is one of the browsers which doesn't accept this and removes the second form. Using a form in a form is bad practice and I suggest you find a different solution to do the form handling.

Slightly left-of-field answer, but have a look at CodeIgniter Form Generator. I've used it a couple of times and it seems pretty good for generating forms from an array. It's a bit tricky to get your head around it to begin with, but it works well once you've gotten into it.
The basic idea is that you implement a form controller from your ordinary controller, and then just output it in your view file. It might be a more elegant (and sustainable) solution to what you're trying.

Related

Two Search bars on one page in magento

I am working on one website where I want to integrate Magento default search two times on a page. I have two search fields and want to search from two stores. Is that possible ?
I can do that for one store and default Magento also support one store at a time. If someone suggest some hints or guidelines it will be appreciated.
you try the blog file using below code...
<?php
echo $this->getLayout()->createBlock('core/template')->setTemplate('catalogsearch/form.mini.phtml')->toHtml() ?>
?>
You could attempt to change the store view using a querystring parameter in the search action. Replace YOUR_STORE below with the store view code you're wanting to search. This will take you to the search results of the other store.
<form id="search_mini_form" action="/index.php/catalogsearch/result/?___store=YOUR_STORE" method="get">
<div class="form-search">
<label for="search">Search:</label>
<input id="search" type="text" name="q" value="" class="input-text" maxlength="128" autocomplete="off">
<button type="submit" title="Search" class="button"><span><span>Search</span></span></button>
<div id="search_autocomplete" class="search-autocomplete" style="display: none;"></div>
<script type="text/javascript">
//<![CDATA[
var searchForm = new Varien.searchForm('search_mini_form', 'search', 'Search entire store here...');
searchForm.initAutocomplete('/index.php/catalogsearch/ajax/suggest/', 'search_autocomplete');
//]]>
</script>
</div>
</form>
If you wanted to integrate both sets of search results into one search results page then that would require some additional coding.

Multiple form in MVC 3 and Ajax

My page in working perfectly with postback. My problem is that it's kind of anoying every ingredients or subtitles that people submit, it reloads the entire page. So, I though that it could be a good way to learn a little bit of ajax... I've read a lot of article, and I got confused. Some people are using Ajax.Beginform and others are using the $.ajax from jQuery inside an event of jQuery (submit for example). I've read that the second approach is better but I don't know if it's possible with the way my form is done.
Here is the important part of my View. To summarize, I have a list of subtitle, and each of the subtitle can contain a list of ingredients. One of the form can submit ingredients, and the other can submit subtitle. The first one can appear multiple time (in each of the subtitle).
<div id="Ingredients">
<h2>Ingrédients</h2>
#foreach (RecettesMaison.Models.Subtitle sub in Model.Subtitles)
{
<h4>#Html.DisplayFor(modelItem => sub.Name)</h4>
<ul id="ing#nSubtitle">
#foreach (RecettesMaison.Models.Ingredient ing in sub.Ingredients)
{
<li>#Html.DisplayFor(modelItem => ing.QuantityAndName)</li>
}
</ul>
using (Html.BeginForm("AddIngredient", "Recipe", new { subname = sub.Name }, FormMethod.Post))
{
#Html.HiddenFor(model => model.IDRecipe)
<a name="IngredientSection" ></a>
<input class="field required span6 text-box single-line" id="nameingredient" name="nameingredient" />
<input type="submit" class="btn btn-success" value="Ajouter un ingrédient" />
}
nSubtitle++;
}
#using (Html.BeginForm("AddSubtitle", "Recipe", FormMethod.Post))
{
#Html.HiddenFor(model => model.IDRecipe)
<a name="SubtitleSection" ></a>
<input class="field required span6 text-box single-line" name="Name" />
<input type="submit" class="btn btn-success" value="Ajouter une catégorie d'ingrédient" />
}
</div>
My first approach was that if a submit is successful, I would "refresh" the two foreach with the new data. But the approach I've seen on must tutorial is to use a partial view and refresh only the partial view inside of a div. But in my case, my Html.BeginForm would be inside the partial view so I don't think it would work. I also think about just appending html at the end of my list but it will only work of the ingredient. So the best way would be to refresh both for each
So, my general question
How can I do that? :)
Thanks!
Ajax.BeginForm vs .ajax()
Both accomplish the same thing. Ajax.BeginForm will hide the javascript details so .ajax() will probably give you a better foundation in understanding the pattern especially if you ever need to work outside of MS technologies.
AJAX with Partial Views
It's entirely possible and there are many examples on the web even here on SO. The general idea is you're just replacing a div with partial view content retrieved via AJAX.
If the div is static, then your jquery selectors are quite simple.
If the div is dynamically generated, then you'll need to do some DOM traversing to find the target div.
Without seeing your attempts it is hard to give you specific help. I suggest you start with a simple example with a single partial view and form and update it using AJAX. Then you can move to a more complex situation with multiple forms once you understand the mechanics.

joomla module development with form - how to process

I'm creating a simple Joomla 2.5 module that will have an html form.
mod_mymodule/tmpl/default.php:
<form method="post" id="myemailform" action="">
<label for="ReferralName">Enter Name:</label><input type="text" name="name" value="<?php echo modCamcloudReferralHelper::getReferralName(); ?>">
<label for="ReferralEmail">Enter Email Address:</label><input type="text" name="email">
<label for="ReferralMessage">Enter Message (optional):</label><textarea class="message"></textarea>
<span class="countdown"></span>
<button type="submit" value="Send Email">Send Email</button>
<?php echo JHtml::_('form.token'); ?>
</form>
I have a helper class at:
mod_mymodule/helper.php - this just has some utility functions in it.
My question is what is the usual convention here to process my form on the server side. I tried to find examples of what people have done but I can't seem to find anything. Do I just put everything in the helper class:
<form method="post" id="myemailform" action="..\helper.php">
Or something like that? Thanks in advance.
Yes, you should do form processing in module helper class. Keep any logic out of the template file, and you can use mod_mymodule.php to call helper methods and assign variables before including the view file.
Do not set as form action helper file! I think in your case action should be the same page, so you can also ommit action url.
Edit: As requested in the comments, this would be the content of your mod_mymodule.php
// include helper file
require_once dirname(__FILE__).'/helper.php';
// call some method of the helper class
$items = modMymoduleHelper::getItems();
$moduleclass_sfx = htmlspecialchars($params->get('moduleclass_sfx'));
// render view file from mod_mymodule/tmpl/default.php, $items is available in the view file
require JModuleHelper::getLayoutPath('mod_mymodule', $params->get('layout', 'default'));

Load Dojo form from ajax call

I am trying to implement something like this.
http://app.maqetta.org/mixloginstatic/LoginWindow.html
I want the login page to load but if you click the signup button then an ajax will replace the login form with the signup form.
I have got this to work using this code
dojo.xhrGet({
// The URL of the request
url: "'.$url.'",
// The success callback with result from server
load: function(newContent) {
dojo.byId("'.$contentNode.'").innerHTML = newContent;
},
// The error handler
error: function() {
// Do nothing -- keep old content there
}
});'
the only problem is the new form just loads up as a normal form, not a dojo form. I have tried to return some script with the phaser but it doesnt do anything.
<div id="loginBox"><div class="instructionBox">Please enter your details below and click <a><strong>signup</strong>
</a> to have an activation email sent to you.</div>
<form enctype="application/x-www-form-urlencoded" class="site-form login-form" action="/user/signup" method="post"><div>
<dt id="emailaddress-label"><label for="emailaddress" class="required">Email address</label></dt>
<dd>
<input 0="Errors" id="emailaddress" name="emailaddress" value="" type="text"></dd>
<dt id="password-label"><label for="password" class="required">Password</label></dt>
<dd>
<input 0="Errors" id="password" name="password" value="" type="password"></dd>
<dt id="captcha-input-label"><label for="captcha-input" class="required">Captcha Code</label></dt>
<dd id="captcha-element">
<img width="200" height="50" alt="" src="/captcha/d7849e6f0b95cad032db35e1a853c8f6.png">
<input type="hidden" name="captcha[id]" value="d7849e6f0b95cad032db35e1a853c8f6" id="captcha-id">
<input type="text" name="captcha[input]" id="captcha-input" value="">
<p class="description">Enter the characters shown into the field.</p></dd>
<dt id="submitButton-label"> </dt><dd id="submitButton-element">
<input id="submitButton" name="submitButton" value="Signup" type="submit"></dd>
<dt id="cancelButton-label"> </dt><dd id="cancelButton-element">
<button name="cancelButton" id="cancelButton" type="button">Cancel</button></dd>
</div></form>
<script type="text/javascript">
$(document).ready(function() {
var widget = dijit.byId("signup");
if (widget) {
widget.destroyRecursive(true);
}
dojo.parser.instantiate([dojo.byId("loginBox")]);
dojo.parser.parse(dojo.byId("loginBox"));
});
</script></div>
any advice on how i can get this to load as a dojo form. by the way i am using Zend_Dojo_Form, if i run the code directly then everything works find but through ajax it doesnt work. thanks.
update
I have discovered that if I load the form in my action and run the __toString() on it it works when i load the form from ajax. It must do preparation in __toString()
Firstly; You need to run the dojo parser on html, for it to accept the data-dojo-type (fka dojoType) attributes, like so:
dojo.parser.parse( dojo.byId("'.$contentNode.'") )
This will of course only instantiate dijits where the dojo type is set to something, for instance (for html5 1.7+ syntax) <form data-dojo-type="dijit.form.Form" action="index.php"> ... <button type="submit" data-dojo-type="dijit.form.Button">Send</button> ... </form>.
So you need to change the ajax contents which is set to innerHTML, so that the parser reckognizes the form of the type dijit.form.Form. That said, I urge people into using a complete set of dijit.form.* Elements as input fields.
In regards to:
$(document).ready(function() {});
This function will never get called. The document, youre adding innerHTML to, was ready perhaps a long time a go.
About Zend in this issue:
Youre most likely rendering the above output form from a Zend_ Dojo type form. If the renderer is set as programmatic, you will see above html a script containing a registry for ID=>dojoType mappings. The behavior when inserting <script> as an innerHTML attribute value, the script is not run under most circumstances (!).
You should try something similar to this pseudo for your form controller:
if request is ajax dojoHelper set layout declarative
else dojoHelper set layout programmatic

Form Submit using a Javascript to invoke webflow transition, doesn't take the updated value on form

I am trying to invoke a form submit using javascript (jquery) to invoke a webflow transition. It works and the submit invokes the desired transition. But, the updated radio button values is not reflected on the model object which is posted.
Here is the code:
<form:form method="post" action="#" commandName="infoModel" name="pageForm">
<form:input type="input" path="testMsg" id="success" />
<input type="button" id="clearSelections" value="Clear Selections">
<div class="question">
<h4><c:out value="${infoModel.questionInfo.description}"/> </h4>
<form:radiobuttons path="infoModel.answerId"
itemValue="answerId" itemLabel="answerDescription" items="${infoModel.answers}" delimiter="<br/>" />
</div>
<input type="submit" name="_eventId_saveQualitativeInput" value="Save" id="save" />
$(document).ready(function() {
$('#tabs').tabs();
//Clear selections (copy is server-side)
$('#clearSelections').click(function() {
//$('input[type="radio"]').prop('checked', false);
$('input[type="radio"]').removeAttr('checked');
$('#save').trigger('click');
});
});
</form:form>
The form:radiobutton, generates the below html:
<div class="question">
<h4>Is this a general obligation of the entity representing a full faith and credit pledge? </h4>
<span>
<input type="radio" checked="checked" value="273" name="infoModel.answerId" id="infoModel.answerId1">
<label for="infoModel.answerId1">Yes</label>
</span>
<span><br>
<input type="radio" value="274" name="infoModel.answerId" id="infoModel.answerId2">
<label for="infoModel.answerId2">No</label>
</span>
<br>
<span class="error"></span>
</div>
The input id= "success" value is registered and when the control goes to the server, the value of input id= "success" is updated in the "infoModel" object. But the value of answerId is not updated on the "infoModel" object.
Thoughts if i am missing something in the form:radiobutton element or if there is something else wrong?
Thanks in advance!
EDIT:::::::
Thanks mico! that makes sense. I stripped of some of the code first time to make it precise, but i have a list which is being used for building the radio-buttons, below is the code:
<c:forEach items="${infoModel.list["index"]}" var="qa" varStatus="rowCount">
<div class="question">
<h4><c:out value="${question.questionInfo.description}"/> </h4>
<form:radiobuttons path="list["index"][${rowCount.index}].answerId" itemValue="answerId" itemLabel="answerDescription" items="${question.answers}" delimiter="<br/>" />
<br>
</div>
</c:forEach>
Could you please suggest how i could try this one out?
NOTE: The same code works on a regular form submit on click of a button of type submit. Its the javascript form submit which is not working. I also tried to do whatever i want to do in javascript and then invoke the button.trigger('click'); form got submitted but the changes made on form in my javascript didnt reflect.
With commandName inside a form:form tag you set "Name of the model attribute under which the form object is exposed" (see Spring Documentation). Then in path you should tell the continuation of the path inside the model attribute.
With this said I would only drop the extra word infoModel from path="infoModel.answerId" and have it rewritten as path="answerId" there under the form:radiobutton.

Resources