I have no code to really show you here, but let's say I have a form with 3 HIDDEN fields that contain date, city and address. I also have a select with 3 options (let's say Apple, Microsoft and Google).
What I want is that when a user changes the select to a different option, jquery should send the value of the selectbox + the 3 hidden fields to a PHP page, say proces.php. Proces.php handles the mysql_query etc, and it doesn't give anything back.
Can anyone show me how this is done? I don't expect anyone to write a whole script for me since I didn't provide any html code, but just the outline, or maybe a link to a tutorial or something.
First of all your form should look something like this:
<form action="process.php" method="post" id="myForm">
<select name="site" id="site">
<option>Apple</option>
<option>Google</option>
<option>Microsoft</option>
</select>
<input type="hidden" name="date" value="01/05/2012" />
<input type="hidden" name="city" value="London" />
<input type="hidden" name="address" value="[... address ...]" />
</form>
Then to submit via AJAX you would use the serialize() method to gather the form data:
$("#site").on("change", function() {
var $form = $("#myForm");
var method = $form.attr("method") ? $form.attr("method").toUpperCase() : "GET";
$.ajax({
url: $form.attr("action"),
data: $form.serialize(),
type: method,
success: function() {
// do stuff with the result, if you want to
}
});
});
Alternatively if you don't want to use AJAX, just the standard form submission, you can trigger the form to submit, like this:
$("#site").on("change", function() {
$("#myForm").submit();
});
if you want to send the form using jquery you can use Jquery Form plugin<
http://jquery.malsup.com/form/
so what basically u can do is to use
$(document).ready(function(){
$('form selector').ajaxForm(opts);
//u can define the opt urself itz easy look at the link
$("selects selector").change(function(){
$(form selector).ajaxSubmit(function(){//something you wanna do if form is submitted successfully})
});
});
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