I am using Roshan's Ajax dropdown code for my two ajax dropdown menus. All is working fine, but when I want to use $_POST in the submitted form with FireFox, its not working...??? Its working fine in IE.
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my issue is both a php and a javascript (AJAX) one its one or the other. I have two combo boxes one dependant on the other, populated by AJAX but when I submit the form the secondary combo is not being registered as a php $_POST['var'] in the following action page. The secondary combo is contained and altered using divs and inner.HTML. Is this an expected behaviour or is my issue else where?
View 3 Replies View RelatedI need to create three dropdown boxes (box 1:colour, box 2: price, box 3:brand) and when the user presses "GO" they are returned with the correct information, pulled from corresponding fields in my MYSQL database.
For example, colour: silver, price £100-£200, brand:samsung...should return 5 items on the page. I do not need the dropdown boxes to populate depending on ther previous selection - they are static, so to speak. If possible, I would like the information to be retuned to my page without the URL changing, which leads me to think I will need to be ausing ajax....?
I want to use prototype js to have it submitted without refreshing the page.
View 4 Replies View RelatedAjax and jQuery is something I'm new to, but PHP, Javascript, and MySQL I'm pretty well seasoned with.For starters, I want to create a simple form where there are these fields:Name (text field)E-mail (text feld)Zip Code (text field)Cell Phone Number (text field)Phone Carrier (dropdown menu)Thing is, I want to check the database to see if an e-mail address has already been entered. I'd like for that to happen after they enter the email address and tab or click out of the e-mail text field. For instance, after they enter an e-mail address and tab to the Zip Code field, the script queries the database to see if that e-mail exists already. If it does, it notifies the user and greys out the submit button. If not, they're allowed to finish filling the form out and submit. It'll be a huge time saver if this Ajax script is in place.
View 1 Replies View RelatedThis form retrieves results from a mysql database. It works fine but when I click submit it shows the result only for a second. When I hold down submit the result stays there. How do I get the result to stay on the screen after I hit submit? Even better how do I get it to spit out results without having to hit a submit button or press enter?
View 15 Replies View RelatedI'm using a script I picked up from elsewhere that allows you to do uploading through AJAX (well, actually through an IFRAME,but lets not get into whether that's really AJAX ). Anyway, the script works great except for one little problem though.Sometimes when it submits the ajax form and gets to the server, my $_POST data is empty.The only correlation I've found is I believe it happens when the file I'm uploading is too big (though I'm not exactly sure what the cut off mark is). Regardless, small files aren't causing problems and large ones mean I have no $_POST data on the server when I get there. Here's the script I'm using:
Code:
/**
*
* AJAX IFRAME METHOD (AIM)[code]....
why this might destroy my POST data?
I have found the following code and wish to have it do a POST instead of the GET
ajax.php
<?php
if (is_numeric($_GET['client_id'])) {
include("database.php");
$query="SELECT * FROM `client_addresses` WHERE `client_id` = '". db_input($_GET['client_id']) ."'";
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I have email, password and some other fields, and I'm using $.post to send data for Ajax submission.
The problem is that I don't want to submit a particular field. The serializeArray() returns all the fields in the form. So, I tried something like this to prevent the password field being serialized.
This works great but the ajax submission doen't work.
I'm created a set amount of dynamic elements based on some elements in another form to try to submit them when the current form is submitted and this just doesnt work. It creates all the elements succesfully and puts in their proper names and values, but on the next page they are not available in $_POST in php.Here's the exact javascript I'm using, I tried appendChild and innerHTML,
Code:
var elem = document.getElementById('submission_options').elements;
for(var k = 0; k < elem.length; k++)
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What i want to do is populate a dropdown with data brought from mysql. However there is are two radio buttons i have which i want to choose what data the dropdown i populated with. This is the code i have so far. I don't have no js yet as my skill is php and mysql, not javascript. Code:
View 3 Replies View RelatedI am trying to create a dropdown that selects mysql with php. For example I have two select dropdowns the first with types and second with products. So when the types select is click the products from that category are listed in the product select.
<select name="type" id="type">
<option value="1">Accessories</option>
<option value="2">Cables</option>
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I'm trying to make 3 drop downs (based off of mysql tables), to narrow down the selection. Source, School, and Program. So when when they select SOURCE(source), all the SCHOOLs in drop2 will be based on the SOURCE selected in the previous dropdown, and whatever school(cid) is selected, the program will be based off that school(cid) and display all the programs for that school. I've google'd high and low for something, and no avail. Here is what I have, I am just trying to make 3 drop downs narrow in, without pressing submit (I am new to ajax and javascript etc) The issue may be in that java part, I've look in the php and mysql, but nothing.
PHP Code:
<table width="517" border="1"> <tr>
<td width="87">Source:</td> <td width="414">
<? echo'<select name="source" onChange="showUser(this.value)">
<option value="">--Select--</option><?php while() { } ?>'; .....
My javascript
Code:
var xmlhttp;
function showUser(str){
xmlhttp=GetXmlHttpObject();
if (xmlhttp==null){
alert ("Browser does not support HTTP Request");
return;
} .....
Right now, the results display in a table, I don't want that - I need them to display as results in the next dropdown.
I am having no luck with my post form to mysql. It is only posting the first letter of the word/phrase for the text input fields to the mysql database. It is a photo and text post form. The photo and only first letter of text are posting to database. I am using Jqtouch, jquery, and phonegap.
Here is my form fields:
Here is my java script:
I'm using formwizard, and I tried to insert a dropdown into the form by putting this jquery code
$('#aog').change(function() {
if ($('#aog').val()==='yes') {
$.ajax ({
type: "POST",
url: "<?php echo base_url(); ?>index.php/register/mortgator",
data: "aog=" + $('#aog').val(),
success: function(msg) {
$("#bank").html(msg);
$("#signupForm").formwizard("update_steps");
}}); //end of ajax
} else {
$("#bank").html('');
}});
//end of aog event
And this php code
function mortgator(){
$this->load->model('MBank');
$q = $this->MBank->getbanklist();
echo '<label for="mortgator">Mortgator:</label><br/>';
echo '<select name="mortgator" id="mortgator" class="required ui-wizard-content ui-helper-reset ui-state-default error">';
echo '<option value="">Select a Mortgator</option>';
foreach ($q as $key => $list){
echo '<option value="'.$list['bankID'].'">'.$list['bankName'].'</option>';
} echo '</select>';
}
It is working, except that when you click next, other form input boxes would fade, but this input inserted via ajax would not have a fade animation. That's why I added this line of code to the html tag ui-wizard-content ui-helper-reset ui-state-default error but it appears that that does not work.
This one should be easy. In fact, it was working for a long time as far as I know. Then I upgraded from jQuery 1.4.4 to 1.7.1 and a few things stopped working. (This is according to my client; it's possible it's been broken for a long time.)
So a customer logs in and sees his/her registration information. Name, address, phone number, etc... It gets this information from a MySQL database. For the customer's home address, there's a dropdown for the state (MA, NH, etc...) Rather than displaying the default specified in the HTML code, I have a bit of jQuery code that is supposed to change the dropdown selection to the user's home state as stored in the MySQL database.
Unfortunately, nothing happens. It rather than changing to the user's home state, it just displays the default state (AL, the first alphabetically).
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I verified that $row['state'] contains valid data by outputting it as HTML on the page (my test data outputs "MA"). And I'm reasonably sure that this used to work in the past.
I have a mysql database (name) which helds 3 fields, id, naam, email. In php I've made the connection and create a selection menu and with javascript I was able to display, (when I selected for example "Bas" in the selection menu), bas his name into a first input text field. Now a couple of things I still want but can't seem to get it right.
First: I also want to display "bas" his email address in a second input text field. (the database contains more names, but this an example).
Second: The selection menu starts with the first name in the database, but can I make it for example "Select name" as beginning.
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i have a form with fields (30+), and about 5 of them are fields that need to pass into my dropdown (actually a dynmaic dropdown, select SOURCE, then whatever source u select it'll show options that are mysql source=$source), so it can go into my mysql query, and filter out the best results. (i.e date_of_birth, min_credits, state, etc). Needs to be done without submitting, hence javascript. Im echo'ing my query and its saying the variables i'm trying to pass are UNDEFINED. Am I not passing the vars correctly? do i need to prep the vars to "grab" them in the current field?
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I want to populate a combobox with data pulled from a mysql db . The value of second combobox depends on what's been selected in first combobox. I assume pulling the data completely when loading the combobox and then filtering it locally to populate the second combobox based on the selection in first combobox will suit my needs so that i do not have to query server.What is the best way of doing it?
View 3 Replies View RelatedI am trying to create a form that writes text to an HTML canvas when submitted. Eventually, the function that writes the text will be more complex. The problem is the text only appears briefly, because the function is only called once when the form is submitted. I want the function to be called continuously after the form is submitted.How do I do this? I have had very little experience with JS.A lame (failed) attempt...
<html>
<head>
</head>[code]......
I am re-developing my website and want to use some Ajax/jQuery in it (so I am learning as I go) and I am not sure if this is possible so that is my first question:I have a registration form and I would like to know if after the person registering has completed their first and surname and the 2 sections of their postcode if I can use jQuery to run a PHP script to see if they are already registered before waiting for the form to be submitted and running a script then.I am thinking onkeyup (or similar) after the last field is completed but I don't know how that would work needing to also use 3 other field values.If this is possible, can anyone give me any ideas of examples that do this or how I can go about passing all the variables and running the scripts at the correct time.
View 6 Replies View RelatedVery intermittently, I am finding ajax requests submitted with jquery are being submitted twice, once with parameters, and once without parameters. The code looks something like this:
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how do you get a textbox in a form to retain an inputed value after that form is submitted
View 4 Replies View RelatedAm creating a framed chat application and when the user types a message in the form field and clicks the submit button, the message gets sent to the display frame, but the message stays in the form field. How can i get the form to submit the message AND reset the form field to blank too?
View 1 Replies View RelatedI have a form,
It has multiple submit buttons (Save/Quit)
I want my form calidation to run when I save and not when I quit.
I still want to submit the form when I quit.
So how do I conditionally run the validation?
i have tried using submit() command to submit the form but the form does not get submitted. I wonder why.I am using lotus designer to develop the form:
var f= document.forms[0];
var answer = confirm('Do you want to referback?')
if(answer ){
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