I have a page which consists of three tabs each with labels. Clicking the label opens up each tab (by showing/hiding the divs). Each tab contains MySQL data, with arrows at the bottom of each tab to go the next 'page' of data.
My Problem, when you click the arrow, the page does indeed go to the next page of data, BUT the default tab (tab1) is made visible and the other two hidden, so if you're on tab two, hit the next arrow, you're taken back to tab1 and have to hit tab3 to see the data.
Link to site here
The script to change visibility of the tabs is here:
<script type="text/javascript">
$(document).ready(function () {
$("div.tab-headers>a").click(function () {
// Grab the href of the header
var _href = $(this).attr("href");
// Remove the first character (i.e. the "#")
_href = _href.substring(1);
// show this tab
tabify(_href);
});
tabify();
});
function tabify(_tab) {
// Hide all the tabs
$(".tab").hide();
// If valid show tab, otherwise show the first one
if (_tab) {
$(".tab a[name=" + _tab + "]").parent().show();
} else {
$(".tab").first().show();
}
}
// On page load...
$(document).ready(function () {
// Show our "default" tab.
// You may wish to grab the current hash value from the URL and display the appropriate one
tabify();
});
</script>
and tab2's code is:
<div class="tab">
<a name="tab2"></a>
<img src="images/glossary_shiptype.png" width="1643" height="952" />
<div class="glossary_body">
<table width="740" border="0">
<?php do { ?>
<tr>
<td><?php echo $row_ship_type['term']; ?>:</td>
<td><?php echo $row_ship_type['definition']; ?></td>
</tr>
<?php } while ($row_ship_type = mysql_fetch_assoc($ship_type)); ?>
</table>
</div>
<div class="glossary_arrow_back">
<?php if ($pageNum_ship_type > 0) { // Show if not first page ?>
<img src="images/arrow_left.png" />
<?php } // Show if not first page ?>
</div>
<div class="glossary_arrow_forward">
<?php if ($pageNum_ship_type < $totalPages_ship_type) { // Show if not last page ?>
<img src="images/arrow_right.png" />
<?php } // Show if not last page ?>
</div>
</div>
hi you have to pass tab id in paramater #tab1 or #tab2 or #tab3 and if you get #tab2 in query string and moving to #tab3 you have to remove #tab2
Related
I have this problem: I can't find out how to hide a div that appears just when session begins. So I a have a X button for closing, and when the page is refreshed, the div appears again! But I don't want to stop the session.
My code:
<div id="got-points" style="display:<?php echo (isset($customer) && !empty($customer) && isset($customer['customer']) && !empty($customer['customer'])) ? 'block' : 'none'; ?>"> // Checks if session is active
<div class="got-points-bg" style="display: visible;">
<div class="got-points-box">
<img class="got-points-close" src="<?php echo base_url();?>static/images/i8.png" /> //Close Button
My text here
</div>
</div>
</div>
And js
<script>
$(document).ready(
function() {
$(".got-points-close").click(function() {
$(".got-points-bg").hide("fast");
});
});
</script>
If it is on refresh then it is easy; just do it in the PHP session; like this:
HTML:
<div id="got-points" style="display:<?php echo (isset($customer) && !empty($customer) && isset($customer['customer']) && !empty($customer['customer']) && empty($_SESSION['first_load'])) ? 'block' : 'none'; ?>"> // Checks if session is active
<div class="got-points-bg" style="display: visible;">
<div class="got-points-box">
<img class="got-points-close" src="<?php echo base_url();?>static/images/i8.png" /> //Close Button
My text here
</div>
</div>
</div>
and in your PHP before then (like on login/session start) - this will obviously need significant editing since you haven't included your PHP I can't guess it very well, but this should provide a starting point:
<?php
//load user check pseudo code
if !empty($_SESSION[user]) and whatever other checks... {
$_SESSION['first_load'] = False; //important line number 2
}
//login pseudo code to demonstrate placement within your code
if user & pass = valid {
$_SESSION['first_load'] = True; //important line number one
}
That is assuming you want th div output but not displayed on subsequent loads.
Interesting question, how about something like this?
<?php
if(isset($_SESSION['variable'])){
sleep(3); /* Duration(in seconds) div is displayed for */
echo
"<script>
$( document ).ready(function() {
$(".got-points-bg").hide("fast");
});
</script>";
}
?>
try this
<?php error_reporting(0); session_start(); if(empty($_SESSION['user'])) echo "<div id='togglediv'> My text here </div>"; ?>
hope it will work
I'm building a website where the admin can make settings for the website. I would like the admin settings page to have a similar "feel" as the rest of the website, which has some nice looking jQuery features.
On the admin site there's a hidden div, which is shown when one of six links has been clicked. I'd like the content of the hidden div to change content before showing itself. I'm not sure how to do this. I could have a div box for every link on the page. But this becomes pretty cumbersome since I'd need to repeat my css and jquery for every link. I imagine that this, somehow, can be done with some javascript/jquery code that determines which link that was click and then decides which php function to call inside the hidden div. Then the php could "echo" out the content of the div, which then could be shown.
How could one do this?
My HTML/jQuery code is as follows:
--- The html links ---
<table id="settings">
<tr>
<td>
<img src="images/folder.gif" alt="" height="100"/>
</td>
<td>
<img src="images/folder.gif" alt="" height="100"/>
</td>
<td>
<img src="images/folder.gif" alt="" height="100"/>
</td>
----- The Hidden div -----
<div id="dashboard_box">
<div class="dashboard_inside">
<form action="#" method="post">
<p style="font-size:20px; font-weight: bold;">Change color</p>
</br>
<fieldset>
<? load_content1();?>
</fieldset>
</form>
</div>
</div>
---- jquery code (working)----
var mouse_is_inside = false;
$(document).ready(function() {
$(".action").click(function() {
var loginBox = $("#dashboard_box");
if (loginBox.is(":visible"))
loginBox.fadeOut("fast");
else
loginBox.fadeIn("fast");
return false;
});
$("#dashboard_box").hover(function(){
mouse_is_inside=true;
}, function(){
mouse_is_inside=false;
});
$("body").click(function(){
if(! mouse_is_inside) $("#dashboard_box").fadeOut("fast");
});
});
You probably want to use ajax to load the content from the server. Take a look at jquery's .load() method: http://api.jquery.com/load/
You could include a data attribute per link:
<a class="action" data-content="content.php?method=load_content1"></a>
<a class="action" data-content="content.php?method=load_content2"></a>
js would look something like this:
$(".action").on('click', function() {
$("#dashboard_box fieldset").load($(this).data('content'), function() {
var loginBox = $("#dashboard_box");
if (loginBox.is(":visible"))
loginBox.fadeOut("fast");
else
loginBox.fadeIn("fast");
}
return false;
});
Then, in your content.php file you could check the method parameter in the url to determine what content to return.
My php is a little rusty, but something like this:
<?
call_user_func($_GET['method']); // i'm not sure how safe this is. you may want to be more explicit
?>
You can just add data attribute in each of your link's
<a href="#" data-url="content1.php" ..
Then on click of any of the a you can get the php to be called.
$('a').on('click',function(){
var phpFunctionToCall = $(this).data('url');
});
You probably need to make ajax call to load content into your fieldset As this <? load_content1();?> run's on server and javascript have no control over it.
Thanks for all the help. this is what I ended up doing.
--- HTML ---
<a class="action" data-content="content.php?method=load_content2"></a>
---Jquery---
var mouse_is_inside = false;
$(document).ready(function() {
$(".action").click(function() {
var phpFunctionToCall = $(this).data('content');
$('#indhold').load(phpFunctionToCall);
var loginBox = $("#dashboard_box");
if (loginBox.is(":visible"))
loginBox.fadeOut("fast");
else
loginBox.fadeIn("fast");
return false;
});
$("#dashboard_box").hover(function(){
mouse_is_inside=true;
}, function(){
mouse_is_inside=false;
});
$("body").click(function(){
if(! mouse_is_inside) $("#dashboard_box").fadeOut("fast");
});
});
--- PHP (shortened)---
function load_settings_panel($settingOnRequest) {
return $settingOnRequest;
}
$result = call_user_func('load_settings_panel', $_GET['method']);
echo($result);
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.
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!
I am trying to make a div where I include PHP web pages into it, however when I click any link inside the div, it loads in new window.
Does anyone know how to solve this problem?
I also tried using an iframe instead of a div but it caused other problems.
HTML code:
<div id="cover5">
<div id="warning">
<?php include 'SignIn.php'; ?>
<p align="center">
Home
</p>
</div>
</div>
JS code:
function cover5() {
var cover = document.getElementById('cover5');
if (vis) {
vis = 0;
cover.style.display='block';
}
else {
vis = 1;
cover.style.display='none';
}
}
You have to make a distinction bitween the client-side and the server-side of a Web application.
<?php include 'SignIn.php'; ?> is a server-side command, because of the <?php ?>, which means that the user will never see that line exactly.
If your SignIn.php file contains for example my link, the user will only get the ouput of that inclusion, i.e.
<div id="warning">
<a href="go-there.php">my link</a
</div>
When the user clicks on the link, his browser will load it as if this <a /> would have been hard written within your HTML code directly, and then open it in a new window.
I don't no how to do it using native JS, but with jQuery, you can try something like this:
$(function() {
$("#warning a").each(function() {
$("#warning").load($(this).attr("href"));
});
});
To render html from a specific link within an element on the page you need to run an ajax GET call, and load the html. Then insert the response in your div.
using jQuery:
$("a").click(function(){ //calls function when anchor is clicked
var link = $(this).prop("src"); //finds the link from the anchor
$.get(link,function(data, status, xhr){ //makes an ajax call to the page
$("#myDiv").html(data); //replaces the html of #myDiv with the returned page
});
});
Hope that helps
HTML CODE :
<div id="cover5">
<div id="links">
<p align="center">
Home
</p>
</div>
<div id="warning">
<?php include 'SignIn.php'; ?>
</div>
</div>
and JS code:
function cover5() {
var cover = document.getElementById('warning');
if (vis) {
vis = 0;
cover.style.display='block';
}
else {
vis = 1;
cover.style.display='none';
}
}
Whenever an user click on the link in "links" div it will call JS function "cover5()"
You can use jQuery or ajax to pop-up "warning" div
In your code you added the links to the worning div so as when in second time ti will hide all the links form there.