Currently I have a PHP file and a .js file. In my PHP file I have div class called wrapper_tab-content. Now the code for this is present in my js file which shows the following code:
$(document).ready(function() {
portalarray = new Array();
$('input.checkbox').change(function(){
portalname = $(this).attr('data-name'); pid= $(this).attr('id');
if ($(this).is(':checked')) {
portalarray.push(pid);
$(".wrapper_tab-content").append('<div class="portalcontent content--active" id="'+pid+'"><div class="col-md-12 text-left">
<label class="control-labels ">Title</label><input id="'+pid+'" name="'+portalname+'" placeholder="'+portalname+' Title" type="text"></div></div>');
}else {
$(".portaltabs .container--tabs li#"+pid).remove();
$(".wrapper_tab-content #"+pid).remove();
tabslength = $(".wrapper_tab-content").length;
}
});
});
Now I have to enter the property title inside value attribute that I can get by using <?php get_portals[0]['property_title'] ?> inside my .js file for which I'm assuming there needs to be an AJAX call being made?
To clear things up for you Javascript is client side and PHP is server side .
What you looking at doing is getting PHP value's into Javascript .There are a few ways to do this . Most programmers use events as server side could have updated during the lifetime of the page
most common is post and get requests through Ajax but you could also get data through an event
<script type='text/javascript'>
document.body.onclick(function(){
var Variable = <?php echo(json_encode($Variable)); ?>;
};
</script>
I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>
You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>
Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.
There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.
Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>
Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);
When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.
I was trying to figure this out myself and then realized that the problem is that this is kind of a backwards way of looking at the situation. Rather than trying to pass things from JavaScript to php, maybe it's best to go the other way around, in most cases. PHP code executes on the server and creates the html code (and possibly java script as well). Then the browser loads the page and executes the html and java script.
It seems like the sensible way to approach situations like this is to use the PHP to create the JavaScript and the html you want and then to use the JavaScript in the page to do whatever PHP can't do. It seems like this would give you the benefits of both PHP and JavaScript in a fairly simple and straight forward way.
One thing I've done that gives the appearance of passing things to PHP from your page on the fly is using the html image tag to call on PHP code. Something like this:
<img src="pic.php">
The PHP code in pic.php would actually create html code before your web page was even loaded, but that html code is basically called upon on the fly. The php code here can be used to create a picture on your page, but it can have any commands you like besides that in it. Maybe it changes the contents of some files on your server, etc. The upside of this is that the php code can be executed from html and I assume JavaScript, but the down side is that the only output it can put on your page is an image. You also have the option of passing variables to the php code through parameters in the url. Page counters will use this technique in many cases.
PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.
We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>
Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>
You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>
Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.
May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.
This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.
We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.
Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.
I've got a basic like button concept on my site that visits url.tld?action=love and adds +1 to the link's database column.
It's a hassle redirecting to another page all the time though. Is it possible to click the button, and send a request to the URL without actually redirecting to a new URL? Also maybe refresh the button afterwards only so that the count updates?
For a general idea of what my download button is this is in the header:
<?php require_once('phpcount.php'); ?>
<p class="hidden"><?php
$time = time();
for($i = 0; $i < 1; $i++)
{
PHPCount::AddHit("$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]", "127.0.0.1");
}
echo (time() - $time);
/*echo "PAGE1 NON: " . PHPCount::GetHits("page1") . "\nPAGE1 UNIQUE: " . PHPCount::GetHits("page1", true);
echo "\n\n" . PHPCount::GetHits("page2");
$ntot = PHPCount::GetTotalHits();
$utot = PHPcount::GetTotalHits(true);
echo "###$ntot!!!!$utot";*/?></p>
And this is an example of my "love" button.
Love <span class="count">'. PHPCount::GetHits("$package_get?action=love", true).'</span>
The reason I used this method is because people create pages, and I wanted the like button to work out of the box. When their page is first visited it adds their url to the database, and begins tallying unique hits.
This is basically adding a new link column called downloadlink?action=love, and tallying unique clicks.
use the following code. assgin id="btn_my_love" to that button and add this code to you page
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
//assign url to a variable
var my_url = <?php echo "https://alt.epicmc.us/download.php?link='.strip_tags($package_get).'?action=love"; ?>;
$(function(){
$("#btn_my_love").click(function(){
$.ajax({
url:my_url,
type:'GET',
success:function(data){
//comment the following result after testing
alert("Page visited");
},
error: function (request, status, error) {
alert(request.responseText);
}
});
//prevent button default action that is redirecting
return false;
});
});
</script>
Yes, it is possible. I am assuming you know what ajax is and how to use it, if not I am not going to give you the code because some simple reading on ajax as suggested by #Black0ut will show you how. But the basic steps are as follows:
Send ajax request to a PHP script that will update +1 vote to the database
In the PHP script, add +1 to the database and return some data to the ajax, maybe the new number of votes
Parse the return data in your JavaScript and update the button accordingly
I have a database table which I am trying to retrieve data from using JQUERY AJAX. When my first page loads it does a php call to a table and populates a select form element. - This works
I then want to select one of the options submit the form and have the row returned via Ajax.
Previously I had the script working with just PHP files but am having trouble getting it to work. When submitting the form my URL is changing:
http://localhost/FINTAN/testertester.php?name=Specifics.
I am not getting anything back. In addition when looking at my console I get a jquery not defined
factory (jquery). I can find the line in question in my jquery ui.js. Not sure if this is the issue or my code has caused the issue. I have cleard the firefox cache and due to the fact I have not had a successful AJAX call via jquery method am guessing it my code.
To get the code below I have mixed and matched a book and an online tutorial and many other sources and this is not my first attempt. Ideally I would like to output table row. However just getting a request working and knowing its not a conflict or compatability issue would makeme feel better and not hindered before I start
<script src="jquery/jquery-ui-1.11.2/jquery-ui.js"></script>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#btn").click(function(){
var vname = $("#name").val;
}
}
$.post("addithandle1.php",
{
name:vname};
function(response,status){
alert("recieved data-------*\n\nResponse : " + response
+"\n\nStatus : " + status);
}
}
</script>
</head>
<body>
<?php
include "config.php";
if (mysqli_connect_errno($con))
{
}
else
{
$result = mysqli_query($con, "SELECT * FROM script ");
echo " <Form method='post'> <label>Script :</label> <select id='name' name='name' >";
}
while($row = mysqli_fetch_array($result))
{
echo "<option value = '".$row['scriptname']."'>".$row['scriptname']."</option>";
}
echo "</select>";
echo "<button id='btn' class='btn-search'>Load Script </button></form>";
?>
</body></html>
This is my PHP file that I am trying to retrieve from
<?php
include 'config.php';
$batchtype2 = $_POST['name'];
$batchtype2 = mysqli_real_escape_string($con,$batchtype2);
$sql = "SELECT * FROM script WHERE scriptname = '".$batchtype2."' ";
$result = mysqli_query($con,$sql);
$count=mysqli_num_rows($result);
if($count==0 ){
echo "</br></br></br></br></br></br></br><p> No Matching results found</p>";
}
else{
while($row = mysqli_fetch_array($result)) {
echo '<tr><td>'.$row['scriptname'].'</td></tr>';
echo '<tr><td>'.$row['scripthours'].'</td></tr>';
echo '<tr><td>'.$row['scripttotal'].'</td></tr>';
}
}
mysqli_close($con);
?>
Thanks in advance for any help
By making the following corrections (you have some syntax issues as well as usage issues which should be revealed in your browser's console when you load this page) in your JavaScript/jQuery this will work like you expect -
Make sure to change this line -
var vname = $("#name").val;
to this -
var vname = $("#name").val(); // note the parentheses
in your function -
$(document).ready(function(){
$("#btn").click(function(e){
e.preventDefault(); // prevent the default action of the click
var vname = $("#name").val();
$.post("addithandle1.php", {name:vname}, function(response, status) { // POST instead of GET
// never use alert() for troubleshooting
// output for AJAX must be in the callback for the AJAX function
console.log("recieved data-------*\n\nResponse : " + response +"\n\nStatus : " + status);
$('#table').html(response); // put response in div
});
});
});
Now $_POST['name'] should get populated properly.
To get the table to appear in your requesting page first make sure that your PHP forms the table completely.
Add a div to your requesting page and modify the AJAX call above as shown.
<div id="table"></div>
Now, when you make a request the div on the requesting page will be updated with whatever comes back from the PHP script.
There are a couple of things about your script.
First make sure you write well structured code and that it is nothing in the wrongplace / broken.
You have in the $(document).ready(function(){ only the .click event of the button, but you left the ajax request outside, I imagine you did that so it will also make the ajax request in the first page load
The problem is that now it will only make it in the first page load, but not when you click the button, on clicking button you are only getting the value of name.
I recommend you to try something like this:
<script>
$(document).ready(function() {
// bind button click and load data
$("#btn").click(function(){
loadData();
return false; // prevent browser behaviour of the button that would submit the form
}
// load data for the first time
loadData();
};
function loadData() {
var vname = $("#name").val;
$.post("addithandle1.php", { name:vname }, function(response, status) {
alert("recieved data-------*\n\nResponse : " + response
+"\n\nStatus : " + status);
});
}
</script>
A few notes:
I would recommend always putting jquery code inside $(document).ready since that guarantees that jquery was loaded before running it
By default a form that has a submit button that you click, will get the form submitted by the browser, if you use ajax, you should prevent that behaviour, either on the button click event or on form with onsubmit="return false".
I have some data which will be displayed like this;
foreach ($holidays as $holiday)
{
$resultTable .= "<p>{$holiday->title}" . "<br/>" .
"{$holiday->pubDate}" . "<br>" .
"{$holiday->description}" . "<input type=\"checkbox\" name=\"saveCB\" value=\"3\"/>" . "<br /></p>";
}
Is there an easy way by which when the checkbox is clicked and the data would be added to a mysql table using AJAX?
Regards Darren
Yes you need javascript to do this. It can be done pretty easily though, if you are satisfied with the form submitting, and the page refreshing each time a select box is changed (i.e. check/unchecked). If you can't accept this, you'll have to use ajax. That would be your optimal solution, and easy as ajax is, it is a nice to have in your toolbox for future projects.
That said, you can achieve this by giving your form an id attribute, and paste this javascript just beneath your form (and edit the form id var):
<script type="text/javascript">
var formId = "YOUR FORM ID HERE";
function submitForm(){document.getElementById(formId).submit()}
</script>
Then add the following attribute to each checkbox: onchange="submitForm()".
Again, it is highly recommended to use ajax for this sort of stuff, and if you look into jQuery ajax, you'll be impressed how easy this can be done.
EDIT: What you can do to actually implement this in your existing code (replace it):
<form action="php-file-to-process-form.php" id="your-form-id" method="post">
<?php if(count($holidays)>0): foreach($holidays as $holiday): ?>
<p>
<?php echo $holiday->title; ?>
<br>
<?php echo $holiday->description; ?>
<input type="checkbox" name="saveCB[<?php echo $holiday->id; ?>]" value="<?php echo $holiday->id; ?>">
</p>
<?php endforeach; endif; ?>
</form>
<script type="text/javascript">
var formId = "your-form-id";
function submitForm(){document.getElementById(formId).submit()}
</script>
Please note i rewrote parts of your code. But in this case, assuming your $holiday objects has an "id" property, php-file-to-process-form.php should receive a fairly comprehensible post request.
PHP doesn't have onClick events, you would have to use JavaScript for something like that.. Or make it so you post your values with PHP (using a form), then it would be possible.
To avoid page refreshing with a form submit you'll want to use AJAX. You didn't tag your question as using jquery, but I highly recommend it. Here is a jQuery example of what you want:
$('input[type=checkbox]').click(function() {
if ($(this).is(':checked')) {
var name = $(this).attr('name');
var value = $(this).val();
$.post('/path/to/your/php/code', {name: value}, function(data){
//Handle the result of your POST here with data containing whatever you echo back from PHP.
});
}
});
Note that this puts the same click handler on all your checkboxes which might be the wrong assumption. If you have other checkboxes on your form that you don't want to use with this logic you'd just need to change the jQuery selector from 'input[type=checkbox]' to something more restrictive such as inputs that have a certain css class.