I'm a stuck with the following function:
<script type="text/javascript">
function removeElement($parentDiv, $childDiv){
if (document.getElementById($childDiv)) {
var child = document.getElementById($childDiv);
var parent = document.getElementById($parentDiv);
parent.removeChild($child);
}
}
</script>
x
This function deletes a child element, and its content, which works great client-side! But I am wanting to pass a value to the server, in the same instance, so the content of the element can be deleted from the mysql database too. I have no idea how to do this, so any suggestions will be very appreciated!
Notes: $child, and $parent are strings generated within the php file, that I use to give each element a unique ID.
To make your life easier, use jQuery or similar framework. Here's how you would do it in jQuery:
$(function() {
$('.delete').click(function() {
var link = $(this);
var id = link.attr('id').replace('element_', '');
$.ajax({
url: 'handler.php',
data: {
element: id
},
type: 'post',
success: function() {
link.remove();
// Or link.closest('tr').remove() if you want to remove a table row where this link is
}
});
return false;
});
});
The HTML:
Remove
And handler.php:
mysql_query("DELETE FROM `table` WHERE id = '".mysql_real_escape_string($_POST['element'])."'");
Always remember to escape database input!
If you're a total noob as you said, you probably won't understand all of this so I suggest you read something about jQuery's AJAX capabilities and about overall development using jQuery or similar JavaScript framework.
Lets say I want to delete an entity using a ID
JQUERY - $.post()
This is an easy way to send a simple POST request to a server without having to use the more complex $.ajax function. It allows a single callback function to be specified that will be executed when the request is complete (and only if the response has a successful response code). Jquery post docs
On the server assuming you have an open database connection.
mysql_query("DELETE FROM TABLE WHERE ID = ".$_POST['ID']);
more on mysql_query found here
EDIT:
So the following will only remove the element when the ajax post is complete. Note the first arg is the url to the script that will take the action , second is the data to be sent, in this case the ID post value will be {child.id} and the third is a anon inline callback function that will take action to remove the element client side.
<script type="text/javascript">
function removeElement($parentDiv, $childDiv){
if (document.getElementById($childDiv)) {
var child = document.getElementById($childDiv);
var parent = document.getElementById($parentDiv);
$.post('{URLTOSCRIPT}', 'ID=$child.id',function () { parent.removeChild($child); });
}}
</script>
When you call the function, you'd want to put your PHP variables in tags like so:
<?php echo $parent; ?>
and
<?php echo $child; ?>
In the function definition, you will want to get rid of the PHP style variables and use something like:
function removeElement(parentDiv, childDiv) {
//CODE
}
Related
I have a table containing data read from a MySQL database via PHP. The first column holds all item names. Now, on clicking a td element in the first column of the table would link to a page with more detailed information about the item contained in the td.
Now I came up with the following idea:
$(document).ready(function() {
$('#table td:first-child').click(function() {
$('div.main').animate({
height: "50px"
}, 600);
setTimeout(function() {
$('div.data').fadeIn(1000);
}, 600);
});
});
div.main is the div-container that has the table included. What I want to do now is to slide that container up and fade a new div-container in, right below it, the new container include()s a PHP page which holds a dynamic query (pseudocode, no string escaping, simplified version):
SELECT detail FROM items WHERE items.name = $_GET['name'];
What I couldn't figure out is if and how I can tell the PHP file that is included in the in-fading div-container which item name it has to grab details for, off the database.
Right now I can read the item name via JavaScript/jQuery, but I couldn't figure a way out to pass that value to the PHP file without having to reload the page.
Any ideas or suggestions welcome!
I think what you're looking for is asynchronous JavaScript and XML (AJAX). It sounds intimidating, but fortunately jQuery makes it very easy.
You can call $.ajax() directly, but for most cases, you can use one of the convenience wrappers. In this case, I think $.load() will meet your needs.
So, let's say your PHP file is called detail_ajax.php and it returns the HTML you wish to put in your div (with class data). All you would have to do then is this:
$('div.data').load( '/detail_ajax.php', function(data){
$(this).html(data);
});
If you want to pass data TO detail_ajax.php, you can pass it along this way:
$('div.data').load( '/detail_ajax.php', { 'someField' : 'someValue' },
function(data) {
$(this).html(data);
}
});
In detail_ajax.php, if you examine $_POST['someField'], you will see the value passed in.
You can do this by using ajax. Output your query on a separate page in JSON format then fetch it using jquery ajax
you need to use ajax to do the same thing. create an event like onclick and call a
method on click call ajax set variable in js and pass it to and do as you want,
show data in particular div in response. Hope it will help you.
You are looking for $.ajax(). However, 3 things will need to take place for this to happen as you intend.
First, we need a reference held in the HTML that is generated by the table so we can streamline the server request. When you generate the table, add a unique data-name string to the TD.
<td data-name="<?php echo $row['name']; ?>">
If, for instance, the td's were generated in a foreach loop, where we expect an array to be returned.
Now, we need to detect the request on our page so we can properly return the data to the browser, we'll look for $_GET['name'] as per your example.
<?php
if(isset($_GET['name'])):
$mysqli = new mysqli('host', 'user', 'pass', 'db');
$ret;
if($stmt = $mysqli->prepare('SELECT detail FROM items WHERE items.name = ?')):
$stmt ->bind_param('s', $_GET['name']);
$stmt ->execute();
$stmt ->bind_result($details); // we only want one column
$stmt ->fetch(); //get our row
$ret['success'] = TRUE;
$ret['html'] = '<div>'. $details .'</div>';
else:
$ret['success'] = FALSE;
endif;
echo json_encode($ret); //return to the browser
endif;
?>
Now we need to employ ajax to bridge the gap between the server and the browser.
Edit - I forgot to modify the click function.
$('#table td:first-child').click(function() {
$('div.main').animate({
height:'0px'
}, function(){
//once the animation completes
$.ajax({
url: '/',
type: 'GET', //this is default anyway
data:{name: $(this).data('name')}, //send the name from the td clicked
dataType: 'json', //what we expect back from the server
success: function(data){ //will fire when complete. data is the servers response
if(data.success !== false){
$('div').html(data.html);
$('div.main').animate({
height: "50px"
}, 600);
}else{
alert("Something went wrong");
}
}
});
}, 600);
});
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($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!
.net developer trying to do a php site for a friend, so far everything is going great but I was wondering if php has something like a textchanged event. Here is what I want to do, I want a drop down box to be appended with data retrieved from a database based on what the user enters in a textbox above(Using the text in the textbox as a parameter to retrieve data from a database and append it to the drop down without reloading the entire page.)
protected void TextBox1_TextChanged(object sender, EventArgs e)
{
//Do stuff
}
The block off code above is in asp.net but i want to implement something similar in php.
That is not how php works. however you can make a ajax call like this with jquery:
<?php
//array, object or db result you use to fill your dropdown
$array = array('pipo', 'kees', 'klaas', 'klaas1', 'jan', 'meneerje', 'poep', 'hessel', 'kaas', 'ietsandersd', 'smit', 'cowoy', 'nog zo iets');
//if we want to search we search and only return the new found options
if(isset($_REQUEST['keyword'])){
$new_array = array();
foreach($array as $value){
if(strpos($value, $_REQUEST['keyword']) !== false){
$new_array[] = $value;
}
}
}
else{
$new_array = $array;
}
$options = '';
foreach($new_array as $key => $option){
$options .= "<option value='$key'>$option</option>";
}
$selectbox = "<select name='selectbox' id='drop_down'>$options</select>";
if(isset($_REQUEST['keyword'])){
echo $options;
}
else{
// with the \ we escape the "
echo "<html>
<head>
<title>ajax selectbox</title>
<script src=\"http://code.jquery.com/jquery-latest.min.js\" type=\"text/javascript\"></script>
<script type=\"text/javascript\">
$(document).ready(function () {
$('body').on('keyup', '.search', function(){
var data = $('.search').serialize();
$.post('ajax_selectbox.php', data, function (data){
$('#drop_down').html(data);
});
});
});
</script>
</head>
<body>
<input type='text' name='keyword' class='search' />
$selectbox
</body>
</html>
";
}
?>
explanation:
java script,
first we include the online jquery library, you can also download the library and include it from your own web server.
<script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script>
<script type="text/javascript">
// first we wait unit the html page is loaded
$(document).ready(function () {
//then we wait for a keyup event in the element with class="search" we use the css sector . for classes like .search
$('body').on('keyup', '.search', function(){
//when we type inside the .search textbox we serialize the element like a form would do. this takes the name and the value and puts it in a array.
var data = $('.search').serialize();
// then we post with ajax back to our php file or an other php file. its you own decision. the data variable is the serialized data form .search
$.post('ajax_selectbox.php', data, function (data){
// at least we use a calback for when the ajax event has finnest and we use the jquery html function to put the new options inside the drobbox with id="drop_down". we use the css id selector # to select the select box.
$('#drop_down').html(data);
});
});
});
</script>
note that I use jquery (and a lot of large players on the web use jquery) and if you know a little java-script the syntax can be disturbing.
In jquery we have a large set of methots we can use directly like:
$.post();
if you want to use the returned data from that function we create a calback function like:
$.post( function(param_ returned_by_parent_function){
//do stuf
});
An other way of using jquery and this is actually the idea behind it is query to a html element and then do stuff with it like this.
$('html_element_query').do_something_with_this();
of course this is just a basic basically explanation but maybe you get the idea.
You can use javascript onChange handler and send the current value to php via AJAX
https://developer.mozilla.org/en/docs/DOM/element.onchange
http://www.w3schools.com/ajax/
PHP does not know what happens on the client. If you want some events on the client to trigger actions, you have to code that yourself (usually in JavaScript).
PHP itself has no awareness of events happening on the front end. You can, however, plug the functionality (kind of) by using a mixture of Ajax and PHP. Ajax will watch for the events and PHP will process data sent to it from that Ajax.
I suggest using jQuery and checking out http://api.jquery.com/Ajax_Events/
I made a very simple PHP Event Dispatcher for myself, it is testable and has been used on my websites. If you need it, you can take a look.
I have a a script that on click do a ajax call connect to the database get imagename and set the image name inside an < -img - > with the right path also it adds a hidden checkbox after it and then echo it.
i then take the ajax message returned and put it as div's HTML. my question is will i be able to preform more action on the inserted content..
The main goal is to be able to click on the image as if it were a checkbox(this part is already sorted for me) however no matter what i try i cant have a .click function works..
Here is the code.
This is the PHP part that echos the images.
if($_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");
});
Hi everyone I have been working on this particular problem for ages by now,plz help.
I have looked at jQuery: Refresh div after another jquery action?
and it does exactly what I want but only once! I have a table generated from db and when I click on delete it deletes the row and refreshes the div but after which none of my jquery functions will work.
$('#docs td.delete').click(function() {
$("#docs tr.itemDetail").hide();
var i = $(this).parent().attr('id');
$.ajax({
url: "<?php echo site_url('kt_docs/deleteDoc'); ?>",
type: 'POST',
data: 'id=' + i,
success: function(data) {
$("#docs tr.itemDetail").hide();
$("#f1").html(data); // wont work twice
//$("#docs").load(location.href+" #docs>*"); //works once as well
}
});
});
in my body I have
<fieldset class='step' id='f1'>
<?php $this->load->view('profile/docs_table'); ?>
</fieldset>
profile/docs reads data from db. <table id='docs'>....</table>
and my controller:
function deleteDoc() {
$id = $_POST['id'];
$this->load->model('documents_model');
$del = $this->documents_model->deleteDocument($id);
return $this->load->view('docs_table');
}
Thanks in advance!
Are you removing any expressions matching $('#docs td.delete') anywhere? If so, consider using $.live(), which will attach your function to ALL matching elements regardless of current or in the future; e.g.
$('#docs td.delete').live('click', function() {
// Do stuff.
});
http://api.jquery.com/live/
Try using bind() instead of click(). The click() method won't work on dynamically added elements to the DOM, which is probably why it only works the first time and not after you re-populate it with your updated content.
You should just have to replace
$('#docs td.delete').click(function() {
with
$('#docs td.delete').bind('click', function() {
Are you replacing the html elements that have the events on them with the data your getting through ajax? If you end up replacing the td.delete elements, then the new ones won't automatically get the binding.