Programmatically submitting a form while using AjaxForm - forms

I wanted to find a way to upload a single file*, in the background, have it start automatically after file selection, and not require a flash uploader, so I am trying to use two great mechanisms (jQuery.Form and JQuery MultiFile) together. I haven't succeeded, but I'm pretty sure it's because I'm missing something fundamental.
Just using MultiFile, I define the form as follows...
<form id="photoForm" action="image.php" method="post" enctype="multipart/form-data">
The file input button is defined as...
<input id="photoButton" "name="sourceFile" class="photoButton max-1 accept-jpg" type="file">
And the Javascript is...
$('#photoButton').MultiFile({
afterFileSelect: function(){
document.getElementById("photoForm").submit();
}
});
This works perfectly. As soon as the user selects a single file, MultiFile submits the form to the server.
If instead of using MultiFile, as shown above, let's say I include a Submit button along with the JQuery Form plugin defined as follows...
var options = {
success: respondToUpload
};
$('#photoForm').ajaxForm(options);
... this also works perfectly. When the Submit button is clicked, the form is uploaded in the background.
What I don't know how to do is get these two to work together. If I use Javascript to submit the form (as shown in the MultiFile example above), the form is submitted but the JQuery.Form function is not called, so the form does not get submitted in the background.
I thought that maybe I needed to change the form registration as follows...
$('#photoForm').submit(function() {
$('#photoForm').ajaxForm(options);
});
...but that didn't solve the problem. The same is true when I tried .ajaxSubmit instead of .ajaxForm.
What am I missing?
BTW: I know it might sound strange to use MultiFile for single-file uploads, but the idea is that the number of files will be dynamic based on the user's account. So, I'm starting with one but the number changes depending on conditions.

The answer turns out to be embarrassingly simple.
Instead of programmatically submitting using...
document.getElementById("photoForm").submit();
... I used...
$("#photoForm").submit();
Also, since I only need to upload multiple files on occasion, I used a simpler technique...
1) The form is the same as my original...
<form id="photoForm" action="image.php" method="post" enctype="multipart/form-data">
2) The file input field is basically the same...
<input id="photoFile" "name="sourceFile" style="cursor:pointer;" type="file">
3) If the file input field changes, submit is executed...
$("#photoFile").change(function() {
$("#photoForm").submit();
});
4) The AjaxForm listener does its thing...
var options = {
success: respondToUpload
};
$('#photoForm').ajaxForm(options);

Related

Passing form value in joomla

I would like to upgrade one of my joomla 2.5 plugin (self-developed). It is a complex task, but here is this specific issue I couldn't solve. I would like to put an input field with a submit button at my articles (done), and after submitting I want to get it. So simple.
Here is the outline of the code:
The form:
<form action="" method="post">
<input type="text" name="info">
<input type="submit" value="ok">
</form>
The process:
$jinput = JFactory::getApplication()->input;
$foo = $jinput->get('info', '444');
print_r($foo);
Basically it should work, but somehow I don't get the value, always recive the default value '444'. If I change the action to an external php file, and process in php-way, it works.
What I checked so far:
a. change form method to GET. Result: the needed value appears properly in the article's URL, but still print the default value '444' not the value I see in the URL (if the default value isn't set, it doesn't print anything).
b. pass the value to an external .php file, store in session, and echo the session value in the article, but empty again.
Maybe I will force to get the GET values by exploding the $_SERVER["REQUEST_URI"], but I can't sleep until I find out what could be wrong with desired process.
Anyone can help?
UPDATE: maybe important - I use K2 plugin.
So far I could figure out the following:
it is a special case. On my local server the code works fine both with normal joomla articles and K2 component articles.
on my website the code also works fine with normal joomla articles, so it is defenietly a K2 settigns-issue.
there was a chance the problem is related to K2 advanced SEF settings (specifies the url of the K2 item), but it isn't. The problem is on my website, so I used on my localhost-version the SEF settings of the website-version, and I recived the values fine. == not K2 advanced SEF settings problem.
This is the answer for my question: "should find which K2 setting is causing the problem"
UPDATE. Solution: turn off caching at global configuration, so the page isn't loading from cache. In cache are no given values stored - obbbbbviously.

Lock/unlock form with toggle

I'm looking for a way to have a button toggeling the field of my form. When "locking" the form with the toggle button no data can be typed. When "unlocking" data should be allowed to be typed. What I want to achieve with this is simple avoiding users to accidentally type.
I found the code below and it works. Only problem is that it only applies to one input field. I want it to work on more that one.
<input type="checkbox" id="yourBox">
<input type="text" id="yourText">
<script>
document.getElementById('yourBox').onchange = function() {
document.getElementById('yourText').disabled = this.checked;
};
</script>
Mark the fields you want to disable with a CSS class, and then use jQuery to disable them.
jQuery - Disable Form Fields
If you want a pure Javascript solution, just repeat this line
document.getElementById('yourText').disabled = this.checked;
for each field.
Or, you can do something like this this: How to Get Element By Class in JavaScript?. Note that you can assign multiple CSS classes to the same field, so assign another class to identify those fields that need to be disabled.

Validation Forms that were loaded via Ajax into jquery-ui tabs

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()
}
});

question about CodeIgniter urls

I am using an application (a blog) written using the CodeIgniter framework and would like to search my blog from my browsers location bar by adding a string to the end of my blogs url like this:
http://mysite.com/blog/index.php/search...
As you can see in the example above I am not really sure how to format the rest of the url after the search part so I am hoping someone here might be able to point me in the right direction.
This is what the form looks like for the search box if that helps at all.
form class="searchform" action="http://mysite.com/blog/index.php/search" method="post">
<input id="searchtext" class="search_input" type="text" value="" name="searchtext">
<input type="submit" value="Search" name="Search">
</form>
Thx,
Mark
Since your form is posting to http://mysite.com/blog/index.php/search, I'm assuming this 'search' controller's default function is the one your are attempting to submit your data to. I think that the easiest way to do this would be to just grab the post data inside of the controller method you're posting to. Example:
function search()
{
$search_params = $this->input->post('search_text');
}
Then you would have whatever the user input stored as $search_params, and you can take actions from there. Am I misunderstanding what you're asking?
It seems like you're kind of discussing two different approaches. If you wanted to make a request to
mysite.com/blog/index.php/search&q=what_I_am_looking_for
This is going to call the search controllers default method (which is index by default). If you wanted to use the URL to pass parameters like that you would go to your function in the search controller and do:
print_r($this->input->get('q'));
This will print out "what_am_I_looking_for".
An easier approach in my opinion would be to:
1. Create a view called "search_view" with the HTML content you pasted above, and have the form "action" http://www.mysite.com/blog/index.php/test/search
Create a controller called "Test" that looks like the following:
class Test extends CI_Controller {
function search()
{
$search = $this->input->post('searchtext');
print_r($search);
}
public function display_search()
{
$this->load->view('search_view');
}
}
Visit http://www.mysite.com/blog/index.php/test/display_search in your browser. This should present you with the form you placed in search_view.php. Once the form is submitted, you should be sent to the search function and print out the variable $search, which will have whatever text you submitted on that form.
If this isn't what you were looking for then I am afraid I do not understand your question.

How can I submit a DOM table to php?

I have a <table> in a web page that I populate using JavaScript code like this:
var cellBarcode = row.insertCell(1);
var textNode = document.createTextNode(barcode);
cellBarcode.appendChild(textNode);
var cellItemName = row.insertCell(2);
var textNode = document.createTextNode(itemName);
cellItemName.appendChild(textNode);
I need to save its data to my database... So I need to know how I can submit it to php... Is it possible...? If yes, please provide some sample codes that are easy to understand for a beginner like me... thanks...
Yes, you can submit this information to a server-side script, but not without extra JavaScript code.
The extra JavaScript code would collect the information in the table (either from the DOM you've built, or by accessing the same data you used when building the table) into a string, which can then be sent to the server using one of standard ways.
Since you've used the term "submit", I'm assuming you want to send the table's data as part of an HTML <form> submission, you can put the generated string in an <input type="hidden"> element.