I have a slider gallery it contains links below the image. I want to
show a popup box with a form and the name of the image to be auto
selected in the popup form.
I have done till this
the popup box is appearing on the link clicked. I have also set the
link with value like this :
<a href="#?id=1" data-reveal-id="myModal" data-animation="fade" >Make an enquiry</a>
This is the code through which i am fetching the value of the link
<?php
if (isset($_GET['id'])){
if($_GET['id']=='1')
echo '<td><input type="text" name="product" style="width:200px; height:30px; !important" value="Testvalue"></td></tr>';
}
else{
echo '<td><input type="text" name="product" style="width:200px; height:30px; !important" ></td></tr>';
}
?>
But I am getting a blank textbox .. Where am I going wrong ???
It should be
Related
I have one doubt is it possible to have a single form(view page) with multiple actions like i want to save,update and delete on the same view page..if the user click on anyof the button then it have to call on a necessary controller function is it possible??
You have some ways to do it, but all of the them require some javascript code.
The easiest I can think of is to dynamically change the form action when clicking each button (of type button, not submit which is the default), and then submit the form.
Example:
<form id="myform" name="myform" method="post" action="">
<input id="myinput" name="myinput" type="text"/>
[..]other inputs[/..]
<button type="button" onClick="deleteAction()">DELETE</button>
<button type="button" onClick="updateAction()">UPDATE</button>
<button type="button" onClick="saveAction()">SAVE</button>
</form>
Where the JS functions are:
function deleteAction() {
changeActionAndSubmit('/action/delete');
}
function updateAction() {
changeActionAndSubmit('/action/update');
}
function saveAction() {
changeActionAndSubmit('/action/save');
}
function changeActionAndSubmit(action) {
document.getElementById('myform').action = action;
document.getElementById('myform').submit();
}
Hope I got your doubt and that this solves your issue :)
A non-JS way to achieve the same goal would be to use the name/value parameters on each button to have your backend decide what to do.
Example
<?php echo form_open('controller/method'); ?>
// form fields go here
<button type="submit" name="add" value="y">press to add</button>
<button type="submit" name="update" value="y">press to update</button>
<button type="submit" name="delete" value="y">press to delete</button>
<?php echo form_close(); ?>
then, on your controller, after validating user input, you can determine which button was pressed by reading what the buttons send about themselves to the controller (I'll assume you use CI's form helper)
if ($this->input->post('add') == 'y')
{
// the user wants to add
}
else if ($this->input->post('update') == 'y')
{
// user wants to update
}
else
{
// user wants to delete
}
// rest of code goes here
on each if structure, you can take the appropriate actions depending on which button the user pressed
I have placed a form (built with entityform) into a modal (ctools modal) using jquery.
Now, what happens is that when I click on the "submit" button, I get redirected to another page where I display the "success" message.
I'd like to avoid the redirect and display the "success" message within the modal (in the same page where the modal pop up).
Thanks in advance.
This is the jquery I used to launch the modal on click:
(function ($) {
Drupal.behaviors.tuo = {
attach: function (context, settings) {
$('#comments_button', context).click(function () {
$('#suggerimenti').dialog('open');
$('.ui-dialog-titlebar').append($('h2', '#block-views-invio-suggerimenti-block'));
$('.ui-dialog-content').append($('form', '#block-views-invio-suggerimenti-block'));
});
$('#suggerimenti').dialog({
autoOpen:false,
minWidth:500,
});
}
};
})(jQuery);
And this is the html of the form created by Drupal:
<form id="prova-invio-entityform-edit-form" class="entityform entitytype-prova_invio-form" accept-charset="UTF-8" method="post" action="/tuo_tema_dev2/?q=content/blandit-ratis-usitas-valde">
<div>
<div class="pre-intructions"></div>
<div id="edit-field-suggerimento-new" class="field-type-text-long field-name-field-suggerimento-new field-widget-text-textarea form-wrapper">
<div id="edit-field-refer" class="field-type-entityreference field-name-field-refer field-widget-entityreference-autocomplete form-wrapper">
<div id="edit-actions--3" class="form-actions form-wrapper">
<input id="edit-submit--4" class="form-submit ajax-processed" type="submit" value="invia suggerimento" name="op">
</div>
</div>
</form>
You can add a hidden form value to your form when using the modal view (check if your ajax is set in the URL). Then in your submit function remove your redirect when the form value is present. Assuming it is set using $form_state using this line:
unset($form_state['redirect']);
We can give you more help if you attach your form code. That seems more relevant to me, for this problem, than your JavaScript.
I have HTML similar to this :
<div class="MainForm">
<form name="FromName">
<button name="Button1"></button>
...
...
</form>
<Div class="blackBox" style="visibility:hidden;"></div>
<Div class="SubFotm" style="visibility:hidden;"></div>
</div>
Now I can properly find the trigger for my button click in my script, but I'm not able to target only the closet blackbox to turn it visible.
Currently I'm doing :
if (PButtonName=="Fermer") {
$(this).closest("div .ProfileForm").remove(); // Closing Profile Form
} else if (PButtonName=="plusAdresse") {
alert('In');
$(this).closest("div .BlackBox").css("visibility","visible");
}
I can get the alert "In" to show, but not the BlackBox
If I change the
$(this).closest("div .BlackBox").css("visibility","visible");
for :
$("div .FormBlackBox").css("visibility","visible");
It will show, but will also show all the black box in the document.
If you are using the above HTML, or something similar, I would do it using a reference to the parents.
instead of:
$(".MainForm").closest("div .BlackBox").css('visibility','visible');
use
$(this).parents('.MainForm').children('.BlackBox').css('visibility','visible');
This is assuming you have more than one MainForm div and they all have a single child with the BlackBox class.
here is an example.
Instead of what you have done just add styles display:none; to your divs and then show them whenever you want.So you can do this as below:
<div class="MainForm">
<form name="FromName">
<button name="Button1"></button>
...
...
</form>
<div class="blackBox" style="display:none;"></div>
<div class="SubFotm" style="display:none;"></div>
</div>
and then in your script
if (PButtonName=="Fermer")
{
$(".MainForm").closest("div .ProfileForm").remove(); // Closing Profile Form
}
else if (PButtonName=="plusAdresse")
{
alert('In');
$(".MainForm").closest("div .BlackBox").show();
}
And I will recommend you using Switch case instead of loops at this place.
I am trying to update records with a image button. here is my code:
<?php
if(isset($_POST['unstar'])) {
echo "good!";
}
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' />
</form>
When I click on the image, it does not echo "good". what am I doing wrong?
I just tested your code and apparently when you click on the image button, php generates the following post values unstar_x and unstar_y, both of which correspond to the x and y coordinates of where you clicked the button.
Try changing it to the following and you'll see what I mean:
<?php
print_r($_POST);
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' />
</form>
EDIT
So if you just need to see if the button was pressed to submit the form, you can just do the following, which just checks if unstar_x or unstar_y was set.
<?php
print_r($_POST);
if(isset($_POST['unstar_x']) || isset($_POST['unstar_y'])){
echo "Good";
}
?>
<form action='' method='post'>
<input type='image' src='http://mysite.com/images/starIcon.png' name='unstar' id='unstar' value="testing" />
</form>
You should note that the output of $_POST will not include the value Testing. At least on my system, I don't get a value when I submit the image button.
The problem is that you're not giving unstar a value, so it won't be detected as Set by the function.
My policyInfoAction redirects my form to clientInfoAction. It stores the empty field errors and then validates the fields in the session variables and redirects it to the client-info page if it contains errors.
It works fine. But the problem is the next time I hit the /client-info page in a new tab it shows the form values in the fields. I have to hit the refresh page to clear it out. I do not want it to display cached data when I open the link in a new tab. What should I do?
public function clientInfoAction(){
//If there are some errors and some valid fields, display the valid fields
$client=$this->session->client;
$state=$this->session->state;
unset($this->session->client, $this->session->state); // delete from the
// assign the values to the view
$this->view->client = $client;
$this->view->state = $state;
}
Here is my view:
<form action ="/pdp/policy-info/" method='post'">
<label for="client_name">Client Name: </label>
<input type="text" name="client_name" id="client_name">
<?php if (!empty($this->client_error)) echo "<font size='2' color ='#C11B17'>".$this->client_error."</font>"; ?>
<br><br>
<label for="state">State: </label>
<select name="state" id='state'>
<option id='state' value="" selected="selected"></option>