Cascading select - php

Sorry in advance everyone for this question as I know the cascading select boxes has been done to death but I can't seem to find any good help. I've tried various things but it all seems to fail and I'm not understanding why.
Here's the jquery I have currently:
function tester() {
$("select#type").attr('disabled', 'disabled');
$("select#cat").change(function(){
var vid = $("select#cat option:selected").attr('value');
var request = $.ajax({
url: "show_type.php",
type: "POST",
data: {id : vid}
});
request.done(function(msg) {
$("#result").html( msg );
});
request.fail(function(jqXHR, textStatus) {
alert( "Request failed: " + textStatus );
});
});
}
Don't mind the first section of the code with the select#type and select#cat as these are for what I was trying to get the code to populate at first, however the .change is my trigger for the .ajax request. The rest of the code I'm merely trying to dump a simple return message into an empty div#result upon a successful ajax request.
I ran a test, and the var vid populates correctly.
Here's the simple PHP file I'm trying to call with the ajax:
<?php
$requ;
if (isset($_POST['id'])) {
$requ = 'Worked';
} else {
$requ = "didn't work";
}
echo $requ;
?>
I thought perhaps the problem was the id wasn't being passed properly so I altered the PHP script to give me any valid output regardless of whether the $_POST was set or not.
I won't post the HTML as I'm just trying to dump this all into a div while I test it. When I run the script I get the 'Request Failed' error message with a message of "error".
Here is the other jquery & PHP I have also tried, using the .post instead of the .ajax:
function tester() {
$("select#type").attr('disabled', 'disabled');
$("select#cat").change(function(){
$("select#type").html("<option>wait...</option>");
var vid = $("select#cat option:selected").attr('value');
$.post("show_type.php", {id:vid}, function(data){
$("#result").empty().append(data);
}, "json");
});
}
And the PHP to accompany this particular jquery:
$requ = $_POST['id'];
$ret = 'You selected: ' . $requ;
echo json_encode($ret);
Again, it all failed. I also tried the above code without using the json encoding/parameters. All I want to do is a simple (so I would think) cascading select dropboxes. The second box to be dependent of the first boxes selection. I'm beginning to think that this all just may not be worth it and just sticking strictly to PHP with links to resubmit the page with a GET and populate a new section or div with the results of the first click. Any help or suggestions you might have would be greatly appreciated, I've spent 2 solid days trying to figure this all out. Thanks in advance

Alright, I got it fixed. Thanks to Mian_Khurram_ljaz for making me take a different look at the hierarchical structure of the file. I was assuming that since the js was calling the php file, by placing the php file in the same folder as the js, I could call the php by using the url: show_type.php but that was actually wrong. The structure is considered from the actual page invoking the js and php, and therefore the url should have been js/show_type.php since I had the show_type.php file in my js folder.
It's always the little mistakes that take you days to figure. For those in the future looking to find decent code for cascading select drop boxes, here is my functioning and fully expanded code (which also includes a tri-level cascade)
jQuery:
function project() {
$("select#model2").attr('disabled', 'disabled');
$("select#brand2").attr('disabled', 'disabled');
$("select#project").change(function(){
$("select#model2").attr('disabled', 'disabled'); // if changed after last element has been selected, will reset last boxes choice to default
$("select#model2").html('<option selected="selected">Choose...</option>');
$("select#brand2").html("<option>Please wait...</option>");
var pid = $("select#project option:selected").attr('value');
$.post("handler/project.php", {id:pid}, function(data){
$("select#brand2").removeAttr("disabled");
$("select#brand2").html(data);
});
});
$("select#brand2").change(function(){
$("select#model2").html("<option>Please wait...</option>");
var bid = $("select#brand2 option:selected").attr('value');
var pid = $("select#project option:selected").attr('value');
$.post("handler/projBrand.php", {proj: pid, bran: bid}, function(data){
$("select#model2").removeAttr("disabled");
$("select#model2").html(data);
});
});
}
Just call the function in the $(document).ready of your js.
Notice the comment, having this 'redundant' call to disable and force the last box to select the default is just in case the user makes a selection in all 3 boxes but goes back to the first box and changes the selection.
Here is the php handler file:
<?php
include_once('../includes/file.inc');
$request = $opt -> getModelvBrand();
echo $request;
?>
The other handler file for the jQuery is nearly exactly the same, only invoking a different method in the class file.
And lastly, the HTML:
<form action="" method="post">
<select id="project">
<option value="0">Choose...</option>
<?php echo $opt -> getProject();?> //populates first box on page load
</select>
<select id="brand2">
<option value="0">Choose...</option>
</select>
<select id="model2">
<option value="0">Choose...</option>
</select>
<br /><br />
<input class="like-button" type="submit" title="Submit" value="" />
</form>
Thanks again Mian for making me take a different look at my file(s).
Hope this code helps someone else in the near future.

Related

Ajax in CI not working

This is the first time I am working with CI+AJAX+JSON.
I have a controller Admin.php inside controllers directory of Codeigniter. The controller code is as under:
public function getmylist()
{
$users_arr[] = array("sno" => "1", "myname" => "hello");
echo json_encode($users_arr);
}
Then, in views, I have a view with the following code:
<select class="form-control" name="mod_countries" id="mod_countries">
<?php
if(isset($countrylist))
foreach($countrylist as $c)
echo "<option value=" . $c->cname . ">".$c->cname;
?>
</select>
Then my target select which i need to populate from ajax is
<select class="form-control" name="mod_newlist" id="mod_newlist">
</select>
My ajax is as under:
$(document).ready(function() {
$('#mod_countries').change(function(event) {
var cname = $("select#mod_countries").val();
alert(cname);
$.ajax({
type: "post",
url: "<?php echo base_url(); ?>" + "Admin/getmylist",
dataType: "json",
data: {mcname: cname},
success: function(res){
if(res) {
var len=res.length;
$("#mod_newlist").empty();
for(var i=0; i<len; i++)
{
var sno=res[i]['sno'];
var myname=res[i]['myname'];
$("#mod_newlist").append("<option value='" + sno + "'>"+myname+"</option>");
}
}
else
{
console.log('hitting');
}
},
error: function(res, status, error) {
var err = res.responseText;
alert(res.Message);
alert(status);
alert(error);
}
});
});
});
When I select an item from the first dropdown list, I get the selected item. Now I expect the second drop downlist to get an item from the controller's code. but, my code goes into error block and alert shows following error messages for each alert line of code:
undefined
parsererror
SyntaxError: Unexpected token I in JSON at position 0
Please tell me where I am making a mistake?
Also, please help with regards to how should I return the sno->name pair from model because at lat, i want to populate the new select from database.
Remove debug/garbage data from response, like I am called. If still not work then parse JSON in success as following
var res = $.parseJOSN(res);
if(res) {
.....
You are not passing data correctly.
it should be like
data:{
"key" : value
//use " " to encapsulate string values, only in case of variable You pass value without ""
}
I am using CodeIgniter for last two months now. And earlier I had a lot of trouble working with ajax. But my friend suggested me to use AngularJs. It took me just one day to learn Angular and it turned out to be extremely helpful.
It will save a lot of time. And make your work so easy. At first, it may sound difficult but if u spend some time coding then it is going to make your life a lot better. I learned it from JavaTpoint.com and it was very well explained.
Do not try to understand everything at once and practice as you learn.

Ajax serialize() form error

There have been similar questions, but I have browsed them a lot and found no accurate answer or fix so please offer a solution!
I am ajaxing a form to a page, and expecting a value back - no big deal. I've done it a million times before, but now it just refuses to work for this one form.
form html:
<form class="selectClaimType" action="place.php" method="post">
<select id="claimtype" name="claimtype">
<option value="privatebuilding">Private Building</option>
<option value="communalbuilding">Communal Building</option>
<option value="outside">Stored Outside</option>
</select>
<input type="submit" id="submit2" name="submit2" class="button" value="save"/>
</form>
jQuery:
$('form.selectClaimType').on('submit',function(e) {
console.log('found');
$form = $(this);
console.log($form.serialize());
e.preventDefault();
$.ajax({
url: "place.php", //$form.attr('action'),
type: "post", //$form.attr('method'),
data: $form.serialize(),
success: function(data) {
console.log(data);
if (data == 1) {
console.log('hello');
}
else {
console.log('failure to change claim type'+data);
}
},
data: function(data) {
console.log('error ajaxing'+data);
}
});
});
The form is not dynamically created, and as you can see I have console.log(ged) nigh on everything. So I know that the form.serialize() is working (values appear as expected). I left out the preventDefault() to test, and the get values were correct.
I have tried dataTypes of script, html, text, xml and json - no success.
I have a var_dump of $_REQUEST and $_POST on the posted to page - these are both empty arrays. I have changed the page that the post is sent to - still doesn't work.
Any ideas at all?
Maybe it is failing. Have you added a fail() callback to see if you get any output from that. Maybe the error is where you are submitting.
data: $form.serialize(),
....
data: function(data) {
console.log('error ajaxing'+data);
}
Probably a conflict here...maybe meant error: ?
This does not answer you question but maybe it solves your problem.
new FormData(form)
is a new native way to send a form.(including files)
Check the pure javascript way to send forms with xhr2.
support all modern browser including mobile devices & ie10
function ajax(a,b,e,d,c){
c=new XMLHttpRequest;
c.onload=b;
c.open(e||'get',a);
c.send(d||null)
}
// Url,callback,method,formdata or {key:val},placeholder
Send the whole Form
var form=document.getElementsById('myForm');
form.onsubmit=function(e){
e.preventDefault();
ajax('submit.php',SUCCESSFunction,'post',new FormData(this));
}
More info
https://stackoverflow.com/a/20476893/2450730 with working copy & past php page.
if you have any questions just ask.

AJAX\JQUERY: Update MYSQL database with form data without refreshing

Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($row = mysql_fetch_array($result))
{
$entry = $row['Entry'];
$CID = $row['CID'];
$checked =$row['Checked'];
// echo $CID;
echo "<input type=\"text\" value=\"$entry\" name=\"textfield$CID;\" id=\"textfield$CID;\" onchange=\"showUser(this.value)\" />";
echo "<input type=\"checkbox\" value=\"\" name=\"checkbox$CID;\" id=\"checkbox$CID;\" value=\"$checked\"".(($checked == '1')? ' checked="checked"' : '')." />";
echo "<br>";
}
?>
<div id="dynamicInput"></div>
<input type="submit" id="checklistSubmit" name="checklistSubmit" class="checklist-submit"> <input type="button" id="CompleteAll" name="CompleteAll" value="Check All" onclick="javascript:checkAll('checklist', true);"><input type="button" id="UncheckAll" name="UncheckAll" value="Uncheck All" onclick="javascript:checkAll('checklist', false);">
<input type="button" value="Add another text input" onClick="addInput('dynamicInput');"></form>
It is populated from the database based on the users session_id, however if the user wants to create a new list item (or is a new visitor period) he can click the button "Add another text input" and a new form element will generate.
All updates to the database need to be done through AJAX\JQUERY and not through a post which will refresh the page.
I really need help on this one. Getting my head around this kind of... Updating method kind of hurts!
Thanks.
You will need to catch the click of the button. And make sure you stop propagation.
$('checklistSubmit').click(function(e) {
$(e).stopPropagation();
$.post({
url: 'checklist.php'
data: $('#checklist').serialize(),
dataType: 'html'
success: function(data, status, jqXHR) {
$('div.successmessage').html(data);
//your success callback function
}
error: function() {
//your error callback function
}
});
});
That's just something I worked up off the top of my head. Should give you the basic idea. I'd be happy to elaborate more if need be.
Check out jQuery's documentation of $.post for all the nitty gritty details.
http://api.jquery.com/jQuery.post/
Edit:
I changed it to use jquery's serialize method. Forgot about it originally.
More Elaboration:
Basically when the submit button is clicked it will call the function specified. You want to do a stop propagation so that the form will not submit by bubbling up the DOM and doing a normal submit.
The $.post is a shorthand version of $.ajax({ type: 'post'});
So all you do is specify the url you want to post to, pass the form data and in php it will come in just like any other request. So then you process the POST data, save your changes in the database or whatever else and send back JSON data as I have it specified. You could also send back HTML or XML. jQuery's documentation shows the possible datatypes.
In your success function will get back data as the first parameter. So whatever you specified as the data type coming back you simply use it how you need to. So let's say you wanted to return some html as a success message. All you would need to do is take the data in the success function and place it where you wanted to in the DOM with .append() or something like that.
Clear as mud?
You need two scripts here: one that runs the AJAX (better to use a framework, jQuery is one of the easiest for me) and a PHP script that gets the Post data and does the database update.
I'm not going to give you a full source (because this is not the place for that), but a guide. In jQuery you can do something like this:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() { // DOM is ready
$("form#checklist").submit(function(evt) {
evt.preventDefault(); // Avoid the "submit" to work, we'll do this manually
var data = new Array();
var dynamicInputs = $("input,select", $(this)); // All inputs and selects in the scope of "$(this)" (the form)
dynamicInputs.each(function() {
// Here "$(this)" is every input and select
var object_name = $(this).attr('name');
var object_value = $(this).attr('value');
data[object_name] = object_value; // Add to an associative array
});
// Now data is fully populated, now we can send it to the PHP
// Documentation: http://api.jquery.com/jQuery.post/
$.post("http://localhost/script.php", data, function(response) {
alert('The PHP returned: ' + response);
});
});
});
</script>
Then take the values from $_POST in PHP (or any other webserver scripting engine) and do your thing to update the DB. Change the URL and the data array to your needs.
Remember that data can be like this: { input1 : value1, input2 : value2 } and the PHP will get something like $_POST['input1'] = value1 and $_POST['input2'] = value2.
This is how i post form data using jquery
$.ajax({
url: 'http://example.com',
type: 'GET',
data: $('#checklist').serialize(),
cache: false,
}).done(function (response) {
/* It worked */
}).fail(function () {
/* It didnt worked */
});
Hope this helps, let me know how you get on!

Jquery script not showing in firebug or firing

I'm trying to get a drop down box to alter a second drop down box through the use of a jquery/ajax script. Firebug is showing Jquery is working but my script isn't showing at all.
<script type="text/javascript">
function ajaxfunction(parent)
{
$.ajax({
url: '../functions/process.php?parent=' + parent;
success: function(data) {
$("#sub").html(data);
}
});
}
</script>
process.php is just a MySQL query (which works)
My initial drop down box is populated by a MySQL query
<select name="front-size" onchange="ajaxfunction(this.value)">
//Query
</select>
And then the second drop down box is just
<select name = "front-finish" id="sub">
</select>
How can I solve this?
calling inline function is not good at all... as web 2.0 standards suggest using unobtrusive JS rather than onevent attributes....check out why here..
other thigs..correct way to use ajax is by using type and data ajax option to send values in controller..
<script type="text/javascript">
$(function(){
$('select[name="front-size"').change(function()
{
$.ajax({
url: '../functions/process.php',
type:'get',
data:{'value' : $(this).val()},
dataType:"html", //<--- here this will take the response as html...
success: function(data) {
$("#sub").html(data);
}
});
});
});
</script>
and your proces.php should be..
<?php
//db query ...thn get the value u wanted..
//loop through it..
$optVal .= "<option value="someDbValue">some DDB values</option>";
// end loop
echo $optValue;exit;
updated
looks like you still have onchange="ajaxfunction(this.value)" this in your select remove that it is not needed and the ajaxfunction in javascript too...
<select name="front-size" >
//----^ here remove that
use jQuery.on() that will allow us to add events on dynamically loaded content.
$('select[name^="front-"]').on('change',function(e){
e.preventDefault();
var value = $(this).val();
ajaxfunction(value);
});
[name^="front-"] this will select all the SELECT box having name starts with front-.
In your process.php give like this
echo "<select name='front-finish' id='sub' onchange='ajaxfunction(this.value)'>";
like this you need to add the "onchange" function to the newly created select box through ajax
or you can remove onchange function and write like
$("select[name^='front-']").live('change',function(){
//Do your ajax call here
});

jQuery, Ajax & PHP submit multiple forms dilemma

This is a very simple form that I have found on the web (as I am a jQuery beginner).
<!-- this is my jquery -->
<script>
$(document).ready(function(){
$("form#submit_wall").submit(function() {
var message_wall = $('#message_wall').attr('value');
var id = $('#id').attr('value');
$.ajax({
type: "POST",
url: "index.php?leht=pildid",
data:"message_wall="+ message_wall + "&id="+ id,
cache: false,
success: function(){
$("ul#wall").prepend(""+message_wall+"", ""+id+"");
$("ul#wall li:first").fadeIn();
alert("Thank you for your comment!");
}
});
return false;
});
});
</script>
<!-- this is my HTML+PHP -->
some PHP ...
while($row_pilt = mysql_fetch_assoc($select_pilt)){
print
<form id="submit_wall">
<label for="message_wall">Share your message on the Wall</label>
<input type="text" id="message_wall" />
<input type="hidden" id="id" value="'.(int)$row_pilt['id'].'">
<button type="submit">Post to wall</button>
</form>
and down below is my PHP script that
writes to mySQL.
It is a pretty straight forward script. However, it is getting little complicated when I submit it. Since I have more than one form on my page (per WHILE PHP LOOP), thus when I submit - only the FIRST form gets submitted. Furthermore, any other subsequent forms that I submit - data is being copied from the first form.
Is there any jQuery functions that clear the data? - or is there a better solution.
Thanks,
Nick
It's because you're giving each form the same id, and thus it is submitting the first element it finds with that id, i.e. the first form. What you should do is assign a unique id to each form, and then give each form an AJAX submit function that submits the form-specific data. You can use jQuery's $.each() function to loop through all the forms and $(this).attr('id') within the submit function to retrieve the form-specific id.
UPDATE: As revealed by the comment on this answer, you actually don't need the each() function because jQuery applies it to every form element anyway.
Here would be an example script:
$(document).ready(function(){
$("form").submit(function() {
var message_wall = $(this).children('input[type="text"]').attr('value');
var id = $(this).children('input[type="hidden"]').attr('value');
$.ajax({
type: "POST",
url: "index.php?leht=pildid",
data:"message_wall="+ message_wall + "&id="+ id,
cache: false,
success: function(){
$("ul#wall").prepend(""+message_wall+"", ""+id+"");
$("ul#wall li:first").fadeIn();
alert("Thank you for your comment!");
}
});
return false;
});
});
Because we can't see all of your forms, I'm not entirely sure, but given your question I'm going to assume that the other forms all share the same id (form#submit_wall), which is invalid an id must be unique within the document.
Given that you're going to change the id of the other forms (I'd suggest using a class name of, probably, 'submit_wall', but the specifics are up to you), the jQuery needs to be changed, too. From:
$("form#submit_wall").submit(function() {
To:
$("form.submit_wall").submit(function() { // using the class-name instead of the id.
Now, of course, you run into the same problems of duplicate ids.
So I'd suggest, again, changing the id to a class and changing:
var message_wall = $('#message_wall').attr('value');
var id = $('#id').attr('value');
to:
var message_wall = $(this).find('.#message_wall').attr('value');
var id = $(this).find('.id').attr('value');
Given the mess that you've posted, above, I find it hard to believe that this is all you need. It would definitely be worth posting the full page (or a demo at JS Fiddle or JS Bin) that fully reproduces your code.

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