I'm trying to create a 'print' button to open a new window and display a PHP Variable in it.
The code below is looped through the PHP script as many times as there are tickets, however I can't seem to get the correct number to display when the window opens (the number that displays in the print link is correct - but when the new window is opened it's incorrect).
<script type='text/javascript'>
jQuery(function($) {
$('a.new-window').click(function(){
var recipe = window.open('','PrintWindow','width=600,height=600');
var html = '<html><head><title>Print Your Ticket</title></head><body><div id="myprintticket">' + $('<div />').append($('#ticket').clone()).html() + '</div></body></html>';
recipe.document.open();
recipe.document.write(html);
recipe.document.close();
return false;
});
});
</script>
Print <?php echo $EM_Booking->get_spaces() ?>
<div style="display:none;">
<div id="ticket"><?php echo $EM_Booking->get_spaces() ?>
</div>
</div>
why not try something like this:
<a href="#" class="new-window">Print <?php echo $EM_Booking->get_spaces() ?>
<div class="ticket" style="display:none">
<?php echo $EM_Booking->get_spaces() ?>
</div>
</a>
<script>
jQuery(function ($) {
$('a.new-window').click(function () {
var btn = $(this),
ticket = btn.find('.ticket').html(),
recipe = window.open('','PrintWindow','width=600,height=600'),
html = '<html><head><title>Print Your Ticket</title></head><body><div id="myprintticket">' + ticket + '</div></body></html>';
recipe.document.open();
recipe.document.write(html);
recipe.document.close();
return false;
});
});
</script>
But much better solution would be to give a button a unique ID, and onclick open an existing (php-generated) page from the server passing that ID, e.g. /getBooking.php?id=123 and that page would output whatever's needed.
Related
I have this code:
<div class="menuList">
<li><img src="/images/icon/arena.png" alt="">Arena<span class="green"> (+)</span>
</li>
</div>
And i want to remove <span class="green"> (+)</span> after users click that link.
Anyone can help me (php code)?
As already mentioned, PHP is not the ideal language to do this in. But, if you need to use PHP, here is how you could do it.
Set a session variable on the /arena/ page, like this;
<?php
session_start();
$_SESSSION['visited'] = 1;
?>
Then, use PHP to check for the session variable in your HTML code like this:
<div class="menuList">
<li><a href="/arena/"><img src="/images/icon/arena.png" alt="">Arena
<?PHP
If(isset($_SESSION['visited'])){
echo '<span class="green"> (+)</span>';
}
?>
</a>
</li>
</div>
You will need to add session_start() to the top of the page wherever you are accessing session variables, before you output anything to the page (e.g. DOCTYPE declaration.)
in Jquery you may do this like
<script>
$(function(){
$('.menuList').find('a').click(function(){
$(this).children('.green').remove();
});
});
</script>
You can accomplish this by adding this javascript to your page:
<script>
window.onload = function() {
var a = document.querySelector('.menuList a');
a.onclick = function() {
var span = a.querySelector('.green');
a.removeChild(span);
}
}
</script>
OR if you're using jQuery:
<script>
$(document).ready(function(){
$('.menuList').find('a').click(function(){
$(this).find('.green').remove();
});
});
</script>
How I can show a specific modal depending on a variable in php which is modified in the url.
I'm using as responsive framework Bootstrap.
I have a page containing several modals, each corresponding to a product. I need to make a link from other parts of the site to go to page "products.php" and automatically display the required product.
E.g:
href="products.php?prod=1"
So far, I have only succeed in showing the modal when loading the page, but if I add or change the variable, it has no effect, not working at all.
This is what I have:
At the beginning
<?php
//Variable to choose which modal to show, default = 1
$prod = 1;
?>
Then the modal which has an id for identification.
<div class="modal fade" id="product1" tabindex="-1" role="dialog" aria-hidden="true">
<!-- Content of the modal -->
</div>
Then the script that show the modal when the page loads.
<script type="text/javascript">
$(document).ready(function(){
<?php if ($prod == 1) {
echo "$(document).ready(function () {var a = 'PD1'; alert(a); });";
echo "$('#product1').modal('show');";
}elseif ($prod == 2) {
echo "$(document).ready(function () {var a = 'PD2'; alert(a); });";
echo "$('#product2').modal('show');";
};
?>
});
</script>
Any idea?
You not gathering the value of the product variable from the URI. To accomplish this at the top of the page where you have $prod=1; you can put the following line of code :
$prod = (isset($_GET['prod']))?$_GET['prod']:"1";
This tells the system to get the value of the prod= in your url if it can't find it then use "1" as a default.
This is not the ideal answer, there are quite a few reasons not to access $_GET/$_POST variables in this way without proper filtering. I leave that for you to research.
You are doing opposite, should be
<?php if ($prod == "1") { ?>
<script type="text/javascript">
$(document).ready(function(){
var a = 'PD1';
alert(a);
$('#product1').modal('show');
});
</script>
<?php } elseif ($prod == "2") { ?>
<script type="text/javascript">
$(document).ready(function(){
var a = 'PD2';
alert(a);
$('#product2').modal('show');
});
</script>
<?php } ?>
I have managed to load the new page into the div (thanks to everyone for your help) but it looks pretty bad (got menu bar and logo, but I only wanted the content), so instead I need to load only a div from the new page. I tried a new script but got redirected to the new page. Please help.
<script>
jQuery(document).ready(function() {
jQuery('.stil_link_img a').click(function(){
var x = $(this).attr("href") + "#content_eco";
jQuery("#new_page").load(x);
return false;
});
});
</script>
<div id="pachete">
<?php
$result=mysql_query("SELECT* FROM imagini");
while($data=mysql_fetch_row($result)){
if( ($data[3]==1)&&($data[2]==2) ){ ?>
<div class="stil_link_img">
<img src="upload/<?php echo $data[1];?>">
</div>
<?php }
}?>
</div>
<div id="new_page">
//some content which should be replaced with my loaded page
</div>
jQuery(document).ready(function() {
jQuery('.stil_link_img a').click(function(){
var $pageContent = jQuery('<div/>').load($(this).attr("href"));
jQuery("#new_page").html(jQuery("#content_eco",$pageContent).html());
return false;
});
});
I assume #content_eco is the divisions ID in the new page(the url from href attribute).
or you can load just the content from the url and avoid the link postback as
<script>
jQuery(document).ready(function() {
jQuery('.stil_link_img a').click(function(){
var x = $(this).attr("rel") + " #content_eco";
jQuery("#new_page").load(x);
return false;
});
});
</script>
<div id="pachete">
<?php
$result=mysql_query("SELECT* FROM imagini");
while($data=mysql_fetch_row($result)){
if( ($data[3]==1)&&($data[2]==2) ){ ?>
<div class="stil_link_img">
<img src="upload/<?php echo $data[1];?>">
</div>
<?php }
}?>
</div>
<div id="new_page">
//some content which should be replaced with my loaded page
</div>
Hope this helps you.
I have a function in jquery which displays images on link click and also has to hide the image when the link is clicked again.
This is the code:
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.min.js"></script>
<script>
var $j = jQuery.noConflict();
$j(document).ready(function(){
$j('.links').click( function(){
var linkclicked = this.id;
var url = $j(this).attr('href'),
image = new Image();
image.src = url;
image.onload = function () {
$j('#image-holder'+linkclicked).empty().append(image);
/* The above line shows the image on first click.
Adding slideToggle hides the image again */
};
image.onerror = function () {
$j('#image-holder'+linkclicked).empty()
.html('That image is not available.');
}
$j('#image-holder'+linkclicked).empty().html('Loading...');
return false;
});
});
function loadnow() {
}
</script>
</head>
<body>
<?php $topicid=1; ?>
<a href="myimages/red-brick-wall-texture.jpg" class="links" id="<?php echo $topicid; ?>" >Show image</a>
<div id="<?php echo 'image-holder'.$topicid; ?>"></div>
<?php $topicid=2; ?>
<br><a href="/myimages/gymicon.JPG" class="links" id="<?php echo $topicid; ?>" >Show image</a>
<div id="<?php echo 'image-holder'.$topicid; ?>"></div>
</body>
</html>
Please let me know where to write slideToggle as to enable show/hide function properly. There are two links here which show separate images. They are shown when clicked but the problem is to hide them on next click and make them
visible when clicked again and so on.
The way you have your code here is not really quite right for .slideToggle. You should not put the URL for the image in the href attribute of an <a> element because the link will try to go there on click.
The way .slideToggle works is well documented on the docs page. All you have to do is use the HTML <img> tag, which is made to hold images. Then you $().hide(); them when the page loads so they aren't shown. All you have to do them is call the .slideToggle() function on the link click event.
See the working example with some misc pictures in this fiddle: http://jsfiddle.net/8xcZT/5/.
Or here is a working version of your code: http://jsfiddle.net/JMXba/9/. I can see this may be desirable to wait for the click to load the image.
Good Luck!
hi i am building a php mysql database pagination page, so i have a list of records 2 rows long at the bottom of the record i want a div which opens up when the span above it is clicked, how do i set up the jquery to make it so that it takes the id of the <p> and expands it in jquery
<span id="button1">Toggle</span>
<p id="p1">
hello<br/>hello
</p>
<script>
$("#button1").click(function () {
$("#p1").slideToggle("slow");
});
</script>
when i output it in php mysql the button will all have a different id and the p will have different ids as well
//Jquery Code which calls toggle_visibility function
$(".pOne").click(function(){
toggle_visibility('partsIdThree');
})
.toggle( function() {
$(this).children("span").text("[-]");
}, function() {
$(this).children("span").text("[+]");
});
//HTML Code
<div id="parts_toogle_one">
<p class="pOne">Add Records <span>[+]</span></p>
<div id="msg_body_one">
<tr><td></td></tr>
</div>
</div>
// Javascript code
function toggle_visibility(id) {
var e = document.getElementById(id);
if(e.style.display == "inline") {
e.style.display = 'none';
}
else if(e.style.display == "none") {
e.style.display = "inline";
}
}
Something like this..
You can use a dynamic id retrieving, so you don't have to worry about the number of results..
For example:
<?php $i = 0;?>
<?php foreach($resultset as $result) : ;?>
<span class="activator" id="result<?php echo $i;?>">CLICK HERE</span>
<div id="panel_result<?php echo $i;?>">SLIDER DIV</div>
<!-- Do the rest of you processing of $result here; it just doesn't matter, for here you only need some identification, I used a number for convenience. -->
<?php ++$i;?>
<?php endforeach;?>
Your jquery:
$('.activator').click(function(){
var the_id = $(this).attr('id');
var the_panel = $('#panel_' + the_id);
$(the_panel).slideToggle('slow');
});
So you don't have to write a jQuery command for each line you print in php, use just this snippet and it will work out by itself which panel to slide, according to which span is clicked.