i am trying to use toggle buttons to save response in db as yes or no. for some reason the only response i am getting is 'on'. even when i switch off the button. i tried searching for problem and got a match but the problem was asked for android platform.now i am stuck with no answer there where similar questions but none of them is useful for me at this moment. sharing the code down below.Thanks in advance for those who are going to suggest or provide a solution.i am using class handicap to save data into variable inside JQUERY and then send that variable to AJAX page to perform db operation.i am not sharing CSS for toggle as i don't think that is required right now. if u need any additional info, do inform me.this input is inside a form with method POST. i am using a submit button with id that is calling this JQUERY.
html part
<div class="switch">
<input id="cmn-toggle-4" class="cmn-toggle cmn-toggle-round-flat handicap" type="checkbox" name="handicap">
<label for="cmn-toggle-4"></label>
</div>
jquery
$("#save-medical-1").click(function () {
var m11 = $(".handicap").val();
alert(m11);
$.ajax({
url: "ajexupdate.php",
type: "POST",
data: {smsgs11: m11},
dataType: 'text',
cache: false,
success: function (e) {
// alert(e);
$("#user_medical_form").html(e);
$("#medidetail").modal('hide');
$('body').removeClass('modal-open');
$('.modal-backdrop').remove();
}
});
return false;
});
You can get value using ":checked" using jquery.
eg.
if($("#cmn-toggle-4").is(":checked")){
m11="yes";
}
else{
m11="no";
}
and send it through ajax.
By writing a php command you are setting the initial value of that input into m11. You have to catch the client side value of input instead:
your code:
var m11 = '<?php echo $_POST['handicap']; ?>'; // always returns the initial value
Correct clien-side code:
var m11 = $(this).val();
Related
Hope someone can help me..
i made my program more simpler so that everybody will understand..
i want my program to get the value of the without submitting, i know that this can only be done by javascript or jquery so I use the onChange, but what I want is when i select an option the value should be passed immediately on the same page but using php..
<select id="id_select" name="name" onChange="name_click()">
<option value="1">one</option>
<option value="2">two</option>
</select>
<script>
function name_click(){
value_select = document.getElementById("id_select").value;
}
</script>
and then i should pass the value_select into php in post method.. i dont know how i will do it.. please help me..
You cannot do this using PHP without submitting the page. PHP code executes on the server before the page is rendered in the browser. When a user then performs any action on the page (e.g. selects an item in a dropdown list), there is no PHP any more. The only way you can get this code into PHP is by submitting the page.
What you can do however is use javascript to get the value - and then fire off an AJAX request to a php script passing the selected value and then deal with the results, e.g.
$(document).ready(function() {
$('#my_select').on('change', do_something);
});
function do_something() {
var selected = $('#my_select').val();
$.ajax({
url: '/you/php/script.php',
type: 'POST',
dataType: 'json',
data: { value: selected },
success: function(data) {
$('#some_div').html(data);
}
});
}
With this code, whenever the selected option changes in the dropdown, a POST request will be fired off to your php script, passing the selected value to it. Then the returned HTML will be set into the div with ID some_div.
not sure ..but i guess ajax is what you need..
<script>
function name_click(){
value_select = $("#id_select").val();
$.post('path/to/your/page',{"value":value_select},function(data){
alert('done')
})
}
</script>
PHP
$value=$_POST['value']; //gives you the value_select data
Post with ajax as Alex G was telling you (+1) and then handle the post with PHP. You can define a callback in Javascript which will run when the page responds.
My suggestion go with jquery. Try with this
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js">
<script>
$(document).ready(function(){
$("#id_select").change(function(){
var url = 'http:\\localhost\getdata.php'; //URL where you want to send data
var postData = {'value' : $(this).value};
$.ajax({
type: 'POST',
url: url,
data : postData,
dataType: 'json',
success: function(data) {
//
},
error: function(e) {
console.log(e.message);
}
});
})
})
</script>
In getdata.php
<?php
var $value = $_POST['value'];
// you can do your logic
?>
I have been going crazy for the last 2 weeks trying to get this to work. I am calling a MySQL Db, and displaying the data in a table. Along the way I am creating href links that DELETE and EDIT the records. The delete pulls an alert and stays on the same page. The EDIT link will POST data then redirect to editDocument.php
Here is my PHP:
<?php
foreach ($query as $row){
$id = $row['document_id'];
echo ('<tr>');
echo ('<td>' . $row [clientName] . '</td>');
echo ('<td>' . $row [documentNum] . '</td>');
echo "<td><a href='**** I NEED CODE HERE ****'>Edit</a>";
echo " / ";
echo "<a href='#' onclick='deleteDocument( {$id} );'>Delete</a></td>";
// this calls Javascript function deleteDocument(id) stays on same page
echo ('</tr>');
} //end foreach
?>
I tried (without success) the AJAX method:
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: 'edit_id='edit_id,
success: function(response){
$('#result').html(response);
}
});
}
</script>
I have been using <? print_r($_POST); ?> on editDocument.php to see if the id has POSTed.
I realize that jQuery/AJAX is what I need to use. I am not sure if I need to use onclick, .bind, .submit, etc.
Here are the parameters for the code I need:
POSTs the $id value: $_POST[id] = $id
Redirects to editDocument.php (where I will use $_POST[id]).
Does not affect other <a> OR any other tags on the page.
I want AJAX to "virtually" create any <form> if needed. I do not
want to put them in my PHP code.
I do not want to use a button.
I do not want to use $_GET.
I don't know what I am missing. I have been searching stackoverflow.com and other sites. I have been trying sample code. I think that I "can't see the forest through the trees." Maybe a different set of eyes. Please help.
Thank you in advance.
UPDATE:
According to Dany Caissy, I don't need to use AJAX. I just need to $_POST[id] = $id; and redirect to editDocument.php. I will then use a query on editDocument.php to create a sticky form.
AJAX is used when you need to communicate with the database without reloading the page because of a certain user action on your site.
In your case, you want to redirect your page, after you modify the database using AJAX, it makes little sense.
What you should do is put your data in a form, your form's action should lead to your EditDocument, and this page will handle your POST/GET parameters and do whatever database interaction that you need to get done.
In short : If ever you think you need to redirect the user after an AJAX call, you don't need AJAX.
You have a SyntaxError: Unexpected identifier in your $.ajax(); request here
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: 'edit_id='edit_id,
success: function(response){
$('#result').html(response);
}
});
}
</script>
it should be like this
<script>
function editDocument(id){
var edit_id = id;
$.ajax({
type: 'POST',
url: 'editDocument.php',
data: {edit_id: edit_id},
success: function(response){
$('#result').html(response);
}
});
}
</script>
note the 'edit_id='edit_id, i changed, well for a start if you wanted it to be a string it would be like this 'edit_id = ' + edit_id but its common to use a object like this {edit_id: edit_id} or {'edit_id': edit_id}
and you could also use a form for the edit button like this
<form action="editDocument.php" method="POST">
<input type="hidden" name="edit_id" value="272727-example" />
<!-- for each data you need use a <input type="hidden" /> -->
<input type="submit" value="Edit" />
</form>
or in Javascript you could do this
document.location = 'editDocument.php?edit_id=' + edit_id;
That will automatically redirect the user
Given your comment, I think you might be looking for something like this:
Edit
$(document).ready(function() {
$('.editLink').click(function(e) {
e.preventDefault();
var $link = $(this);
$('<form/>', { action: 'editdocument.php', method: 'POST' })
.append('<input/>', {type:hidden, value: $link.data('id') })
.appendTo('body')
.submit();
});
});
Now, I don't necessarily agree with this approach. If your user has permission to edit the item with the given id, it shouldn't matter whether they access it directly (like via a bookmark) or by clicking the link on the list. Your desired approach also prevents the user from opening links in new tabs, which I personally find extremely annoying.
Edit - Another idea:
Maybe when the user clicks an edit link, it pops up an edit form with the details of the item to be edited (details retrieved as JSON via ajax if necessary). Not a new page, just something like a jQuery modal over the top of the list page. When the user hits submit, post all of the edited data via ajax, and update the sql database. I think that would be a little more user-friendly method that meets your requirements.
I was facing the same issue with you. I also wanted to redirect to a new page after ajax post.
So what is did was just changed the success: callback to this
success: function(resp) {
document.location.href = newURL; //redirect to the url you want
}
I'm aware that it defies the whole purpose of ajax. But i had to get the value from a couple of select boxes, and instead of a traditional submit button i had a custom anchore link with custom styling in it. So in a hurry i found this to be a viable solution.
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($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!
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.
I'm trying to take values from a dropdown two boxes and send them to a PHP file which will draw an appropriate field from a mySQL database depending on the combination chosen and display it in a div without refreshing the page using AJAX. I have the second part sorted, but I'm stuck on the first part.
Here is the HTML: http://jsfiddle.net/SYrpC/
Here is my Javascript code in the head of the main document:
var mode = $('#mode');
function get() {$.post ('data.php', {name: form.him.value, the_key: #mode.val()},
function(output) {$('#dare').html(output).show();
});
}
My PHP (for testing purposes) is:
$the_key = $_POST['the_key'];
echo $the_key;
After I have it in PHP as a variable I can manipulate it, but I'm having trouble getting it there. Where am I going wrong? Thanks for your replies!
You need a callback function as well to have the server response to the POST.
$.post('ajax/test.html', function(data) {
$('.result').html(data);
});
This snippet will post to ajax/test.html and the anonymous function will be called upon its reply with the parameter data having the response. It then in this anonymous function sets the class with result to have the value of the server response.
Help ? Let me know and we can work through this if you need more information.
Additionally, $.post in jQuery is a short form of
$.ajax({
type: 'POST',
url: url,
data: data,
success: success
dataType: dataType
});
your jquery selectors are wrong:
html:
<select id="mode">
jquery selector:
$("#mode").val();
html:
<select name="player">
jquery selector:
$("select[name=player]").val();
You want to add a callback to your ajax request, its not too hard to do, here ill even give you an example:
$.ajax({
url: "http://stackoverflow.com/users/flair/353790.json", //Location of file
dataType: "josn",//Type of data file holds, text,html,xml,json,jsonp
success : function(json_data) //What to do when the request is complete
{
//use json_data how you wish to.;
},
error : function(_XMLHttpRequest,textStatus, errorThrown)
{
//You fail
},
beforeSend : function(_XMLHttpRequest)
{
//Real custom options here.
}
});
Most of the above callbacks are optional, and in your case i would do the following:
$.ajax({
url: "data.php",
dataType: "text",
data : {name: ('#myform .myinput').val(),the_key: $('#mode').val()},
success : function(value)
{
alert('data.php sent back: ' + value);
}
});
the ones you should always set are url,success and data if needed, please read The Documentation for more information.