placing PHP Variables inside javascript file - php

As per title.
I have a piece of jQuery that looks like the following
$("<div class='creditCardDetails' id='usercurrentccbox'>
<div class='creditCard'>
<div class='cardChoice'>
<span>Choose Card Type</span>
<input name='cctype' type='radio' value='V' class='lft-field' id='visa' />
<label for='visa'></label>
<input name='cctype' type='radio' value='M' class='lft-field' id='mastercard' />
<label for='mastercard'></label><input name='cctype' type='radio' value='A' class='lft-field' id='amex' />
<label for='amex'></label>
</div>
<!--cardChoice-->
<div class='cardNumber'>
<input name='ccn' id='ccn' type='text' class='long-field' value='<?php echo MaskCreditCard($loggedInfo[ccn]);?>' maxlength='19' readonly />
</div>
<div class='cardCVV'>
<input name='cvv' id='cvv' type='text' maxlength='5' class='small-field' />
</div>
<div class='creditCardName'>
<input name='ccname' id='ccname' type='text' class='long-field' value='<?php echo $loggedInfo[ccname];?>' readonly/>
</div
<div class='cardDate'>
<input name='exp1' id='exp1' type='text' maxlength='2' class='small-field' value='<?php echo $loggedInfo[ccm];?>' readonly /><input name='exp2' id='exp2' type='text' maxlength='4' class='small-field' value='<?php echo $loggedInfo[ccy];?>' readonly />
</div>
</div><!--creditCard-->").insertAfter("#paymentCardChoice");
But as you'll see it has PHP Variables, if I have this embedded into my PHP file this works but I want to keep the PHP file as short as possible and placed this code inside a .js file and of course my variables only display the text of the PHP not the variable its self.
The variables I am trying to call are defined already in a config.php file.
Would I need to use something like this? if so I'm stuck with how I would call the variables in the code above.
$.post('phpfile.php', qString, function (data) {
}, "json");

PHP code will not execute in a .js file. A very simple workaround is to put your JavaScript+PHP code in a .php file; and set the content type header, for example:
<?php
#
# script.js.php
#
header("Content-type: text/javascript");
?>
$("<div><?php echo $foo; ?></div>");
You will find the json_encode function very handy for converting just about any PHP variable (string, number, arrays etc) into valid JavaScript. This function takes care of escaping characters such as ", \, \n etc which could break your JavaScript code. Following example demonstrates how you can pass arbitrary PHP variables to your JavaScript code:
<?php
#
# script.js.php
#
header("Content-type: text/javascript");
?>
var config = <?php echo json_encode(array(
"foo" => "foo",
"bar" => "bar",
"baz" => array(
"baz1",
"baz2"
)
)); ?>;
You can include "PHP-powered JavaScript" file in your PHP page using the <script src> tag:
<script type="text/javascript" src="script.js.php"></script>

If you place it in a separate .js file, it will not work since Javascript is client side and PHP is server side.
Your PHP variable only works now since it is declaired earlier in the page before you insert your javascript, so it is rendered when the page loads.

you can have a PHP file which will include your config.php file
include("config.php");
and then you can have your jquery code within script tags
<script>
// you jquery code with php config variables
</script>

.js
var qString = 'selectedID=' +selectedID;
$.post('/assets/inc/get-logged-info-card.php', qString, function (results) {
$("<div class='creditCardDetails' id='usercurrentccbox'>
<div class='creditCard'>
<div class='cardChoice'>
<span>Choose Card Type</span>
<input name='cctype' type='radio' value='V' class='lft-field' id='visa' />
<label for='visa'></label>
<input name='cctype' type='radio' value='M' class='lft-field' id='mastercard' />
<label for='mastercard'></label><input name='cctype' type='radio' value='A' class='lft-field' id='amex' />
<label for='amex'></label>
</div>
<!--cardChoice-->
<div class='cardNumber'>
<input name='ccn' id='ccn' type='text' class='long-field' value='"+results[0].maskccn+"' maxlength='19' readonly />
</div>
<div class='cardCVV'>
<input name='cvv' id='cvv' type='text' maxlength='5' class='small-field' />
</div>
<div class='creditCardName'>
<input name='ccname' id='ccname' type='text' class='long-field' value='"+results[0].ccname+"' readonly/>
</div
<div class='cardDate'>
<input name='exp1' id='exp1' type='text' maxlength='2' class='small-field' value='"+results[0].ccm+"' readonly /><input name='exp2' id='exp2' type='text' maxlength='4' class='small-field' value='"+results[0].ccy+"' readonly />
</div>
</div><!--creditCard-->").insertAfter("#paymentCardChoice");
}, "json");
.php
<?php
$selectedID = $_POST['selectedID'];
$q = "SELECT * FROM tbl_user WHERE user_id = '$selectedID'";
$sql = mysql_query($q);
$results = array();
while($row = mysql_fetch_array($sql))
{
$results[] = array(
'cct' => $row['user_cct'],
'ccn' => $row['user_ccn'],
'maskccn' => MaskCreditCard($row['user_ccn']),
'ccname' => $row['user_ccname'],
'ccm' => $row['user_date_m'],
'ccy' => $row['user_date_y']
);
}
echo json_encode($results);
?>

Related

How start a JQuery inside PHP

I want to start a jquery (freeow plugin) when I got an error in e-mail input field form.
I'm checking some fields on the form and each error that I got I want to open an alert message.
I can do this using a button, like this...
<div id="freeow-tr" class="freeow freeow-top-right"></div>
<div id="freeow-br" class="freeow freeow-bottom-right"></div>
<div class="form-line-check">
<input id="freeow-title" value="Freeow!" type="hidden" class="text" />
<input id="freeow-error" value="0" type="hidden" />
<input id="freeow-position" value="#freeow-tr" type="hidden" />
<input id="freeow-style" value="smokey" type="hidden" />
<input id="freeow-message" value="Error message..." type="hidden" />
</div>
<div class="form-line">
<input id="freeow-show" type="button" value="Click to Freeow!" />
</div>
How do this without the button? Inside the PHP code.
if ($erro == "0") {
if(filter_var($email01, FILTER_VALIDATE_EMAIL)){
$erro = "0";
} else {
echo "<div id=\"freeow-tr\" class=\"freeow freeow-top-right\"></div> ";
echo "<div id=\"freeow-br\" class=\"freeow freeow-bottom-right\"></div> ";
echo "<div class=\"form-line-check\"> ";
echo "<input id=\"freeow-title\" value=\"Freeow!\" type=\"hidden\" class=\"text\" /> ";
echo "<input id=\"freeow-error\" value=\"0\" type=\"hidden\" /> ";
echo "<input id=\"freeow-position\" value=\"#freeow-tr\" type=\"hidden\" /> ";
echo "<input id=\"freeow-style\" value=\"smokey\" type=\"hidden\" /> ";
echo "<input id=\"freeow-message\" value=\"Error message...\" type=\"hidden\" /> ";
echo "</div> ";
$erro = "1";
}
}
You have to create a javascript function
function mensagem01() {
var title, message, opts, container;
title = $("#freeow-title").val();
message = $("#freeow-message").val();
message = 'E-mail not in database...';
opts = {};
opts.classes = [$("#freeow-style").val()];
opts.prepend = false;
opts.classes.push("error");
opts.autoHide = false;
opts.classes.push("slide");
opts.hideStyle = { opacity: 0, left: "400px" };
opts.showStyle = { opacity: 1, left: 0 };
container = $("#freeow-position").val();
$(container).freeow(title, message, opts);
}
HTML code need to have the freeow definitions
<div id="freeow-tr" class="freeow freeow-top-right"></div>
<div id="freeow-br" class="freeow freeow-bottom-right"></div>
<div class="form-line-check">
<input id="freeow-title" value="Lembrou?" type="hidden" class="text" />
<input id="freeow-error" value="0" type="hidden" />
<input id="freeow-position" value="#freeow-tr" type="hidden" />
<input id="freeow-style" value="smokey" type="hidden" />
<input id="freeow-message" value="message field must be empty" type="hidden" />
</div>
Inside PHP code
You need to include the javascript code to call the javascript function message
echo "<script type=\"text/javascript\"> $(document).write(mensagem01()); </script> \n";
When the php is executed the javascript write the document, which contains the call to the freeow message.
What you are trying to achieve will is best done with AJAX and some more logic than you have in your code.
You want to catch the form submit with jQuery. Using event.prefentDefault() you can stop the form from submitting. Then, you want to AJAX your FormData to the server.
In your PHP script, you do your validation. If it succeeds, return something (for example, a JSON object) that tells your front end that the validation was a success. If not, you want an object that describes what exactly is failing.
From front end, you check to see if the validation was a success (do whatever), or not (show freeow)

Modifying $i inside a form

so I have a form. The form consists of 10 lines by default. It goes like this:
<form method="post" action="actionhere">
<?php
for($i=0; $i<10;$i++) {
?>
<div class='clone_me'>
<span>Line <?php echo $i;?></span>
<input type='checkbox' name='ck_<?php echo $i;?>'/>
<input type='text' name='tx_<?php echo $i;?>'/>
</div>
<?php } ?>
<input type='submit' name='submit' value='Submit'/>
</form>
So inside the form, we will have 10 rows of checkbox+textbox.
What I'm trying to make is, I want to place a button to add new row (the checkbox+textbox). Now, problem is, I need the $i value (since it's form the for loop). Is that possible that when we click the add row button, the value of $i that we set inside for loop be incremented by 1 on each click? I know we can clone the div using jquery, but how about the $i value?
I think you are doing it in wrong way you do not need $i value inside name attribute you have to use array for it for example
<form method="post" action="test.php">
<div class='clone_me'>
<span>Line 1</span>
<input type='checkbox' name='ck[]'/><!--this field should menditory-->
<input type='text' name='tx[]'/>
<span>Line 2</span>
<input type='checkbox' name='ck[]'/><!--this field should menditory-->
<input type='text' name='tx[]'/>
</div>
<input type='submit' name='submit' value='Submit'/>
</form>
Now implement this code in actionhere.php
<?php
$cks = $_POST['ck'];
$txs = $_POST['tx'];
foreach($cks as $key => $ck) {
echo $ck."<br>";
echo $txs[$key]."<br>";
}
?>
Well, basically no. Your PHP script is already over when the html has been generated. So you can't rely anymore on PHP. But you don't need to make it explicitly appear in your html.
You should count the rows using jquery :
var i = $('form').find('.clone_me').length
and then add a new row using javascript again :
$('form .clone_me:last').clone().insertAfter('form .clone_me:last');
<input type='hidden' name='counter' id='counter'/>
<input type='checkbox' name='chk' id='chk'/>
<?php
$counter=$_POST['counter'];
for($i=0;$i<=$counter;$i++)
{
$chk=$_POST['chk'.$i];
// Your Insert Code Here
}
?>
may be this can be but have some limit
if you have no problem with page reload the you can do it is:-
by this way your page will reload and the value in textbox and checkbox will gone:---:)
every time new page generate and send by server to client browser
<?php
if(isset($_POST['submit'])){
$ends = $_POST['ttl_rows'];
}else{
$ends = 10;
}
?>
<form method="post" action="#">
<?php
for($i=1 ; $i<$ends;$i++) {
?>
<div class='clone_me'>
<span>Line <?php echo $i;?></span>
<input type='checkbox' name='ck_<?php echo $i;?>'/>
<input type='text' name='tx_<?php echo $i;?>'/>
</div>
<?php } ?>
<input type='hidden' name='ttl_rows' value='<?php echo ($i+1); ?>'/>
<input type='submit' name='submit' value='Submit'/>
</form>

Show/hide div using radio buttons with jquery mobile

Ok here is my problem, and all of my 'dirty' code. This isn't my production code but just trying to make it work at the moment.
Basically what I need is when a user selects the Not Ok radio button it displays the textarea for that unique set which it doesn't do right now when I select Not Ok it gives the textarea's for all the entries which right now is about 13 sets of questions that get generated dynamically from a mysql database at the moment. I have a feeling it has to do something with unique id's that are either in the wrong place in my code now, or just simply aren't there at all. Any help is appreciated greatly.
<div data-role="collapsible" data-theme="b" data-content-theme="c">
<h3>Vehicle Check Information</h3>
<?php
$query = mysql_query("SELECT * FROM vehicle_q");
while($row = mysql_fetch_array($query)) {
$q_title = $row['title'];
$q_id = $row['id'];
?>
<div data-role="fieldcontain" style="border:0;">
<fieldset data-role="controlgroup">
<legend><?php echo $q_title; ?>:</legend>
<input type="radio" name="help[]" id="checkbox-1a" value="Ok" />
<label for="checkbox-1a">Ok</label>
<input type="radio" name="help[]" id="checkbox-2a" value="Not Ok" />
<label for="checkbox-2a">Not Ok</label>
</fieldset>
<div id="hidden_text-<?php echo $q_id; ?>" style="display:none;">
<script>
$(document).ready(function(){
$(":radio:eq(1)").click(function(){
$("#hidden_text-<?php echo $q_id; ?>").show(500);
});
$(":radio:eq(0)").click(function(){
$("#hidden_text-<?php echo $q_id; ?>").hide(500);
});
});
</script>
<fieldset data-role="fieldcontain">
<label for="<?php echo $q_title; ?>_t">Explain the Deficiency(If any):</label>
<textarea name="text_a[]" id="<?php echo $q_title; ?>_t"></textarea>
</fieldset>
</div>
</div>
<input type="hidden" name="q_title1[]" value="<?php echo $q_title; ?>" />
<?php
}
?>
</div>
The other 2 answers are ways to fix your issue. As a side, here is possibly why it is happening. When you are using -
$(":radio:eq(1)").click(function()
$(":radio:eq(0)").click(function()
You are using a click listener that just checks for if 'any' radio button with that index position was clicked on your page. So any button with 1 index position will make every $(":radio:eq(1)").click(function() execute.
Edit -
You would want to change your radio button ids, as (1) they will not be unique as you repeat them with each while() loop, and (2) you could use it to check if that specific radio buttton was clicked.
Try changing it to -
...
<input type="radio" name="help[]" id="checkbox-1a-<?php echo $q_id; ?>" value="Ok" />
<label for="checkbox-1a">Ok</label>
<input type="radio" name="help[]" id="checkbox-2a-<?php echo $q_id; ?>" value="Not Ok" />
<label for="checkbox-2a">Not Ok</label>
</fieldset>
<div id="hidden_text-<?php echo $q_id; ?>" style="display:none;">
<script>
$(document).ready(function(){
$("#checkbox-2a-<?php echo $q_id; ?>").click(function(){
$("#hidden_text-<?php echo $q_id; ?>").show(500);
});
$("#checkbox-1a-<?php echo $q_id; ?>").click(function(){
$("#hidden_text-<?php echo $q_id; ?>").hide(500);
});
});
</script>
The easiest way that I have found to show hide elements based off of radio values is to serve up an onchange event based on the id of the element (which you already have). In your case, since you only have two radios based off the same name, the easiest way is a simple if/else statement. Something like:
<script type="text/javascript">
$(function(){
$('#checkbox-1a').change(function(){
if($(this).attr('checked')){
$('#hidden-text-<?php echo $q_id;?>').hide(500);
}else{
$('#hidden-text-<?php echo $q_id;?>').show(500);
}
});
});
</script>
Where we state when #checkbox-1a is changed: if the element is checked (selected), then do something (in your case hide something), otherwise do something else (in your case show something).
Try something like:
<div data-role="collapsible" data-theme="b" data-content-theme="c">
<h3>Vehicle Check Information</h3>
<?php
$query = mysql_query("SELECT id,title FROM vehicle_q");
while($row = mysql_fetch_array($query)) {
$q_title = $row['title'];
$q_id = $row['id'];
?>
<div data-role="fieldcontain" style="border:0;" class="groupings">
<fieldset data-role="controlgroup">
<legend><?=$q_title?>:</legend>
<input type="radio" name="help[]" id="checkbox-<?=$q_id?>-1a" value="Ok" />
<label for="checkbox-<?=$q_id?>-1a">Ok</label>
<input type="radio" name="help[]" id="checkbox-<?=$q_id?>-2a" value="Not Ok" />
<label for="checkbox-<?=$q_id?>-2a">Not Ok</label>
</fieldset>
<div id="hidden_text-<?=$q_id?>" class="hiddenText" style="display:none;">
<fieldset data-role="fieldcontain">
<label for="<?=$q_title?>_t">Explain the Deficiency(If any):</label>
<textarea name="text_<?=$q_id?>_a[]" id="<?=$q_title?>_t"></textarea>
</fieldset>
</div>
</div>
<input type="hidden" name="q_title1[]" value="<?php echo $q_title; ?>" />
<?php
}
?>
</div>
<script>
$(document).ready(function(){
$(".groupings input:radio").click(function(){
if (this.value == "Ok") {
$(this).parent().next('.hiddenText').show(500);
} else {
$(this).parent().next('.hiddenText').hide(500);
}
});
});
</script>
This is untested, but the script only needs to be declared once and it will apply to the rest relative to the radio button.

How to get selected radio type input and use it as a variable?

How can i get the selected radio input type value in a php script and direct it to a different directory to retrieve the files selected. I tried writing the code, but couldn't write it properly since i m a newbie in the world of php.
The code i have written for displaying radio button is
<input type="radio" name="radio<?php echo $orderby1[0]; ?>"
id="<?php echo $orderby1[0]; ?>" value="<?php echo $orderby1[0]; ?>"
/><?php echo $orderby1[0]; ?>
Now how to get the selected values from it and direct it to the directory contains the file?
The code i have written is , which doesn't seems to work
<script type="text/javascript">
var element = document.getElementById("<?php echo $orderby1[0]; ?>");
jmolApplet(400, "load /jmol/'element.value'.txt");
</script>
I don't want to include any for loops... that makes my script work differently.
Please help!
You would use a $_REQUEST method within a form, something like:
<?php
if ($_REQUEST['radiotest']) {
echo "You chose ".$_REQUEST['radiotest'];
}
?>
<form>
<div><input type='radio' name='radiotest' value='1' /> Option 1</div>
<div><input type='radio' name='radiotest' value='2' /> Option 2</div>
<div><input type='radio' name='radiotest' value='3' /> Option 3</div>
<div><input type='submit' value='Submit' /></div>
</form>
You need first to execute when the radio has rendered and then not put the value inside the quotes
function someEvent() {
var element = document.getElementById("<?php echo $orderby1[0]; ?>");
jmolApplet(400, "load /jmol/"+element.value+".txt");
}

Passing PHP form variables from one page to other

my php configuration on server are showing i can post variables to maximum size upto 8MB , thats enough .. but how to check for number of variables , sever is running ubuntu 4.4 , php .
i have a page which takes students marks and send them to a action page , but on action page doing echo for the post variables nothing is being displayed , where are doing an echo "hello"; this shows ...
this is the page which sends the variables
<form name="frm" action="marklistI.php" method="POST" class="" >
<?php $tb->displayTable() ?>
<div class="mainframe">
<input type="hidden" name="batch" value="<?php print $_GET['batch']; ?>"/>
<input type="hidden" name="sem" value="<?php print $_GET['sem']; ?>" />
<input type="hidden" name="chance" value="<?php print $_GET['chance']; ?>"/>
<input name="submit" type="submit" class="hide" value="Save"/>
<input type="hidden" name="url" value="<?php print $_SERVER['REQUEST_URI']; ?>"/>
</div>
</form>
and this are the variables are coming to action page .. but on echo they are not showing any value .
$dept =$_COOKIE['dept'];
$join=$_POST['batch'];
$type='e';
$sem=$_POST['sem'];
$chance=$_POST['chance'];
try placing this code on your action page:
if (isset($_GET)) {
echo "<h3>GET METHOD</h3>";
var_dump($_GET);
}
if (isset($_POST)) {
echo "<h3>POST METHOD</h3>";
var_dump($_POST);
}
if (isset($_COOKIE)) {
echo "<h3>COOKIE METHOD</h3>";
var_dump($_COOKIE);
}
See which method returns your variables and use it, otherwise, you are not filling any values on the form.
this is your code:
<form name="frm" action="marklistI.php" method="POST" class="" >
<?php $tb->displayTable(); ?>
<div class="mainframe"> <input type="hidden" name="batch" value="<?php print $_GET['batch']; ?>"/>
<input type="hidden" name="sem" value="<?php print $_GET['sem']; ?>" />
<input type="hidden" name="chance" value="<?php print $_GET['chance']; ?>"/>
<input name="submit" type="submit" class="hide" value="Save"/>
<input type="hidden" name="url" value="<?php print $_SERVER['REQUEST_URI']; ?>"/>
</div>
</form>
One possible reason for your issue:
You use "_GET[]" variables here but the form is POST.
GET and POST are two different methods to send data, GET is in the URL path (a=&b=&c=) while POST is hidden in the HTML headers.
So make sure you read those results as "$_POST['name']" and not GET.
I suggest this in the "receiving script" for debugging:
var_dump($_GET);
var_dump($_POST);
And in your browser use Chrome or Firefox + Firebug and Press "f12".
In that debugger you can catch the POST when you click the button and you can look which variables were sent.
That should help you debug your issue fast.
One other suggestion, I personally would write the code less "mixed".
It makes it hard to read and hard to modify.
Why not like this:
<?php
echo "
<form name='frm' action='marklistI.php' method='POST' class='' >".
$tb->displayTable().
"<div class='mainframe'>
<input type='hidden' name='batch' value='$_GET[batch]'/>
<input type='hidden' name='sem' value='$_GET[sem]' />
<input type='hidden' name='chance' value='$_GET[chance]'/>
<input name='submit' type='submit' class='hide' value='Save'/>
<input type='hidden' name='url' value='$_SERVER[REQUEST_URI]'/>
</div>
</form> ";
?>
My guess for your problem is that those values in the formular are actually empty, that's why you don't receive anything.

Categories