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.
When my page opens I call a PHP file and output a form in the form of an echo.
A simplified example below:
echo "<table id='results'>";
echo "<form method = 'post'><input id='".$row['batchname']."'><button class='btn1'>Query this record</button></form>";
</table>
There will be many versions of the above form as table rows are pulled from the database.
I am usin AJAX to handle the output:
$(document).ready(function(){
$(".btn1").click(function(e){
e.preventDefault();
var bname = ("#<?php echo $row['batchname'];?>").val();
$post(
"somephp.php",
{name : bname},
function(response, status){
$("#results").replaceWith(response);
}
);
});
});
When I input an non PHP ID into the jQuery the AJAX work but I always post the first returned row for every form produced as the ids output in the PHP are the same. Can I echo PHP variable into jQuery like this? Is there a better way of getting dynamic ID's into jQuery.
Rather than hacking about like this, do it properly.
$(".btn1").click(function(e) {
e.preventDefault();
var form = $(this).parents("form");
var value = form.find("input")[0].value; // or something else here
});
Ok, so I've gotten most of this thing done.. Now comes, for me, the hard part. This is untreaded territory for me.
How do I update my mysql database, with form data, without having the page refresh? I presume you use AJAX and\or Jquery to do this- but I don't quite grasp the examples being given.
Can anybody please tell me how to perform this task within this context?
So this is my form:
<form name="checklist" id="checklist" class="checklist">
<?php // Loop through query results
while($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!
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');
});
So what I am doing is trying to create a 'favorite' system. I want the user to click a button and the code on the page will submit a value into a MySQL Database. Does this need to the page need to reload if the only thing I am doing is submit and value. I am not pulling any information from the database on the button's click. Thank you:)
A great way to avoid the complexities of Ajax and cross-browser compatibility is to use Jquery!
In your non-reloading page, you can put this:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.7.1.min.js"></script>
<script text="text/javascript">
$(function() {
$('#button_id').click(function() { //in place of "button_id" you need to put your button's id
submitInformation("some information you want to send");
});
});
function submitInformation(data1)
{
$.post(
"handle.php", //this is the name and location of your php page
{
"input_var_one":data1,
}
);
}
</script>
and in your handler php page (in this case called "handle.php")
<?php
$inData1 = $_POST['input_var_one'];
//after your mysql_connect and mysql_select_db
$query = "INSERT INTO `yourtablename` VALUES ('var1 whatever you want', '$inData1')";
mysql_query($query);
?>
Jquery handles the Ajax request for you!
If you do not use AJAX, the information must be sent to the server in order to get to a mysql database table and processed by php and that surely requires page reload.
you can use:
<script type="text/javascript">
var xhr = XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");
function onthatbuttonclick(something)
{
window.xhr.open('GET', "somephp.php?click="+something, false);
window.xhr.send(null);
alert(window.xhr.responseText);
}
var somevar = "user 01";
</script>
<input type="button" onclick="onthatbuttonclick(somevar);" />
and in php:
<?php
// some query required code
// and yes... i does require some safety measures:
$val = mysql_real_escape_string($_GET['click']);
mysql_query("INSERT INTO `tabel` (`click`) VALUES ('".$val."')");
echo 'you cliked a button.';
?>
you should now see an alert box with the text: "you clicked a button.".