I am using editable plugin to edit-in-place. I am able to do it on the web page but I want to change this value to the database also.
Here is the php/html code :
while($row = mysql_fetch_assoc($result))
{
echo "<tr class='highlighter'><td class='editable-1'>".$row['subcategory_name']."</td>";
echo "</tr>";
}
and the jquery code is as follows :
$(document).ready(function()
{
$('.editable-1').editable({onEdit:begin});
function begin(){
this.append('Click somewhere else to submit');
}
}
);
I am trying to use ajax but not getting the value when I am updating in textbox.
Here is what I am trying to do
$.ajax({
url:'change_subcat.php',
data:'NOT AVAILABLE'
});
Please tell me how to made changes also in database.
You will need to submit the data via ajax to the server, where you need a script to process the data and save it to your database.
It looks like in this plugin, you can attach a listener to the onEdit event that will submit your data, maybe something like this:
$('.editable-1').editable({onEdit: submitData});
function submitData (content) {
$.post('/save_to_db.php', { data: content.current });
}
From the documentation for this plugin, it appears that the callback is called with an object for an argument (content, in this case) that has previous and current properties. You should be able to get the new value of the editable field with content.current.
Related
Ok guys. Im new to jquery, and I have a jquery array that I need to pass as a $_POST to the same file which is called index.php. When they click the button I need it to reload the index.php so I can get the POST. Now I'm sure I'm doing something wrong here cause this isn't working. But I'm certain that its something simple that I'm missing, or that I'm doing it all wrong and I mis-understood how this works. Any help is appreciated as I've been on this and trying to figure it out for the last 6 hours.
The index.php also contains the jquery scripts which are below, and the Post check which is
if (isset($_POST['data'])){
echo "ok, data was sent.<br>";
echo " Data is - " .$_POST['data'];
}
And the Button call
echo "<p><input type=\"submit\" class=\"input-button\" id=\"btn-add\" value=\"Add Squad\" /></p>";
Jquery
$(document).ready(function(){
// Get items
function getItems(exampleNr)
{
var count = 0;
var columns = [];
$(exampleNr + ' ul.sortable-list').each(function(){
count++;
columns.push($('#squad'+count).val(), $(this).sortable('toArray').join(','));
});
return columns.join('|');
}
$(document).on('click','#btn-get', function() {
$.post('index.php', {'data': getItems('#squad')}
});
});
You aren't doing anything with the response of your POST.
If you want it to submit like a form, you will need to create a form, attach a hidden input to it with your key-value pair, then submit that form. $.post alone is used for AJAX.
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($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: $('#checklist').serialize(),
cache: false,
}).done(function (response) {
/* It worked */
}).fail(function () {
/* It didnt worked */
});
Hope this helps, let me know how you get on!
I have a a script that on click do a ajax call connect to the database get imagename and set the image name inside an < -img - > with the right path also it adds a hidden checkbox after it and then echo it.
i then take the ajax message returned and put it as div's HTML. my question is will i be able to preform more action on the inserted content..
The main goal is to be able to click on the image as if it were a checkbox(this part is already sorted for me) however no matter what i try i cant have a .click function works..
Here is the code.
This is the PHP part that echos the images.
if($_POST['updateIgallery'] == 'ajax'){
global $wpdb;
$table_name= $wpdb->prefix . "table_T";
$imagecounter = 1;
$toecho = '';
$currentselected = $wpdb->get_row("query");
preg_match_all('/\/(.+?\..+?)\//',$currentselected ['image_gal'],$preresualts); // images are stored with /image/.
foreach ($preresualts[1] as $imagename){
$toecho .= '
<img rel="no" id="JustantestID" class="JustaTestClass" src="'.site_url().'/wp-content/plugins/wp-ecommerce-extender/images/uploads/'.$imagename.'">
<input name="DoorIMGtoDeleteIDcheck'.$imagecounter.'" style="display:none;" name="DoorIMGtoDelete['.$imagecounter.']" value="/'.$imagename.'/" type="checkbox">
';
$imagecounter++;
}
echo $toecho;
}
This is the ajax part that send and receive and insert the HTML to the div:
$.ajax({
type: "POST",
url: "/wp-content/plugins/wp-ecommerce-extender/DB_Functions.php",
data: { updateIgallery: "ajax", CurrentDoorIDnum: $('#dooridforgallery').val()}
}).success(function(insertID) {
$("#ImgGalleryID").html(insertID);
});
This so far works what i am having trouble with is the following:
$("#JustantestID").click(function() {
//DoorImageGallery($(this).attr('id')); // the function i will use if the alert actually works
alert("kahdaskjdj");
return true;
});
I hope the question and the code is understandable.
Thanks in advanced.
When you replace element's html, all the elements inside it are removed and gone. That means the event handlers attached to them are removed as well.
You could try attaching an event handler to a higher level element that is static and permanent on your page. Without more info I am going to use document:
$(document).on( "click", "#yaniv", function() {
alert("kahdaskjdj");
});
$('img.JustaTestClass').bind('click', function() {
var checkbox = $(this).siblings('input[type=checkbox]');
if (!checkbox.is(':checked')) checkbox.attr('checked', true);
else checkbox.attr('checked', false);
});
Since the elements are dynamically inserted into the DOM with ajax, you have to delegate events to a parent element that actually exists when binding the click handler, which in this case looks to be #ImgGalleryID
$('#ImgGalleryID').on('click', '#yaniv', function() {
DoorImageGallery(this.id);
alert("kahdaskjdj");
});
The attached picture shows the results page of the search engine that I'm building. For each return result, the user may click on the result (i.e. "Food Science") and it will expand out accordion-style to reveal information about that particular result.
I want to log each time the user clicks on a result (for learning/intelligence purposes) and store it in a database table that I have created which stores the session ID, the query, the position of the result, and the order in which the user clicked the item.
Using JQuery, I already have a function that will pull the title of the result that was clicked, and I have it set where I want to log the click, but I don't know how to do it since JQuery is client side and PHP is server side.
How can I use the JQuery to trigger a PHP function so that I can query the database to insert the click logs into my table?
Below is the JQuery function.
$(document).ready(function() {
$('.accordionButton').click(function(e) {
if($(this).next().is(':hidden') == true) {
$(this).addClass('on');
$(this).next().slideDown('normal');
$(this).next().slideDown(test_accordion);
// SEND CLICK ACTION TO LOG INTO THE DATABASE
alert($(this).find('h3:last').text()); // displays the title of the result that was just clicked
}
else {
$(this).removeClass('on');
$(this).next().slideUp('normal');
$(this).next().slideUp(test_accordion);
}
});
}
You can do something like this (untested):
Define a javascript variable to track the order of the clicks, outside your click function:
var order = 0;
Add this into your click function, at the bottom:
order++;
var sessionID = $("input[name='sessionID']").val(); // assuming you have sessionID as the value of a hidden input
var query = $("#query").text(); // if 'query' is the id of your searchbox
var pos = $(this).index() + 1; // might have to modify this to get correct index
$.post("logClick.php", {sessionID:sessionID, query:query, pos:pos, order:order});
In your php script called "logClick.php" (in the same directory):
<?php
// GET AJAX POSTED DATA
$str_sessionID = empty($_POST["sessionID"]) ? '' ; $_POST["sessionID"];
$str_query = empty($_POST["query"]) ? '' ; $_POST["query"];
$int_pos = empty($_POST["pos"]) ? 1 ; (int)$_POST["pos"];
$int_order = empty($_POST["order"]) ? 1 ; (int)$_POST["order"];
// CONNECT TO DATABASE
if ($str_sessionID && $str_query) {
require_once "dbconnect.php"; // include the commands used to connect to your database. Should define a variable $con as the mysql connection
// INSERT INTO MYSQL DATABASE TABLE CALLED 'click_logs'
$sql_query = "INSERT INTO click_logs (sessionID, query, pos, order) VALUES ('$str_sessionID', '$str_query', $int_pos, $int_order)";
$res = mysql_query($sql_query, $con);
if (!$res) die('Could not connect: ' . mysql_error());
else echo "Click was logged.";
}
else echo "No data found to log!";
?>
You can add a callback function as a third parameter for the $.post() ajax method if you want to see if errors occured in the script:
$.post("logClick.php", {sessionID:sessionID, query:query, pos:pos, order:order},
function(result) {
$('#result').html(result); // display script output into a div with id='result'
// or just alert(result);
})
);
EDIT: If you need the value of the order variable to persist between page loads because you paginated your results, then you can pas the value of this variable between pages using either GET or POST. You can then save the value in a hidden input and easily read it with jQuery. (Or you could also use cookies).
Example (put this in every results page):
<?php
$order = empty($_POST["order"]) ? $_POST["order"] : "0";
$html="<form id='form_session' action='' name='form_session' method='POST'>
<input type='hidden' name='order' value='$order'>
</form>\n";
echo $html;
?>
In your jQuery, just change var order = 0; to
var order = $("input[name='order']").val();
Then, when a user clicks on a page link, prevent the default link action, set the order value and the form action, and then submit the form using javascript/jQuery:
$("a.next_page").click(function(event) {
event.preventDefault();
var url = $(this).attr("href");
$("input[name='order']").val(order);
$("#form_session").attr('action', url).submit();
});
All the 'next' and 'previous' pagination links must be given the same class (namely 'next_page' (in this example).
EDIT: If your pagination is as follows:
<div class='pagination'>
<ul><li><a href='page1.url'>1</a></li>
<li><a href='page2.url'>2</a></li>
</ul>
</div>
then just change this:
$("div.pagination a").click(function(event) {
etc.
This one is pretty easy, you need a PHP-Script to handle AJAX requests which are sent from your Search page.
In your search page you'll need to add an .ajax to create an AJAX request to your Script.
Everything you need to know about AJAX can be found here: http://api.jquery.com/jQuery.ajax/
In your PHP-Script you'll handle the Database action, use GET or POST data to give the script an ID over Ajax.
Use Ajax. Write a simple php-script that writes clickes to the database. I don't know how you log the clicks in the database exactly, but you can send the clicked item unique identifier to a php script with ajax, for example via POST variables.
A little example, on click:
$.post(
'count_click.php',
{ id: "someid" },
function(data) {
// data = everything the php-script prints out
});
Php:
if (isset($_POST['id'])) {
// add a click in the database with this id
}
Send a request to a PHP page using jQuery AJAX. See here for more info (it is really simple):
http://api.jquery.com/jQuery.ajax/
In this particular case, as you do not need to return anything, it may be better to just use the POST or GET methods in jQuery:
http://api.jquery.com/jQuery.post/
http://api.jquery.com/jQuery.get/
Something like:
$.ajax({
type: "POST",
url: "some.php",
data: "name=John&location=Boston"
success: function(data){
alert('done');
});
ajax is not yet sothin i master.
I have two forms field
code :
name :
and the submit button like :
<form><input type=text name=code><input type =text name=name/></form>
I would like in php/jquery to check if the code the user fill exist in a table of my db.
If it does not exits, when the user leave the textfield to fill the next one, i would like to print a message like: this code is not in the db and then clean the fied. Until the user provide a valide code.
If your php service returns true or false for validation.
and the placeholder for the error is a label called
then an example (in jQuery) would be
$(document).ready(function() {
$("form").submit(function(e) {
var code = $("input[name='code']");
var error = $("#error");
e.preventDefault();
var form = this;
$.getJSON('urlToPhp',
{ code: code.val() },
function(valid) {
if (!valid) {
error.text(code.val() + ' is not found try another code...');
code.val('');
} else {
form.submit();
}
}
);
});
});
I've created a simple example at http://jsfiddle.net/nickywaites/e4rhf/ that will show you have to create a jQuery ajax post request.
I'm not too familiar with php so that part of it I'll have to leave aside although you can use something along the lines of $_POST["Name"].
Here is php example that I googled http://php4every1.com/tutorials/jquery-ajax-tutorial/ that might be better for you.