I have a page in php home.php and it has views something like this
home.php
<?php
?>
<div id="showView">
Show User
Show Car
</div>
<div id="userView">
<p>This is view where user profile is shown</p>
</div>
<div id="carView">
<p>This is view where user car is shown</p>
</div>
<script>
$(document).ready(function()
{
$("#user").click(function(e)
{
//This where I am confused
var car = $("#carView").html();
$("#showView").append(car);
});
});
</script>
Now I am confused that how to pick userView and append to showView I tried a bit of coding but it didn't help and it didn't show result. Moreover is there any way that when that userView is appended then its html is shown when I view-source a page?
You have two options:
1) you generate the html on the fly using jquery (Not recommended)
<div id="showView">
Show User
Show Car
<div id="view_content"></div>
</div>
<script>
$(document).ready(function()
{
$('.view_trigger').click(function () {
var viewHTML = '';
if ($(this).attr('id') == 'user') {
viewHTML = '<div id="userView"><p>This is view where user profile is shown</p></div>';
} else if ($(this).attr('id') == 'car') {
viewHTML = '<div id="userView"><p>This is view where user car is shown</p></div>';
}
$("#view_content").html(viewHTML);
});
});
</script>
2) show/hide the container on trigger
<div id="showView">
Show User
Show Car
</div>
<div id="userView" class="view_content" style="display:none">
<p>This is view where user profile is shown</p>
</div>
<div id="carView" class="view_content" style="display:none">
<p>This is view where user car is shown</p>
</div>
<script>
$(document).ready(function()
{
$('.view_trigger').click(function () {
var viewHTML = '';
$('[class=view_content]').hide();
if ($(this).attr('id') == 'user') {
$('#userView').show();
} else if ($(this).attr('id') == 'car') {
$('#carView').show();
}
});
});
</script>
I don't recommend any of these solutions, you should use MVC for a better code.
Related
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've been trying to learn yii recently and so i started to work on an simple form with bootstrap. im using yii 1.1.x version not version 2 so i had to install bootstrap manually and write coding based on it.
anyway coming back to the problem it seems that my has-success class is being called correctly by jquery but it does'nt seem so for the has-error...
can someone help and im grateful for your help
<script>
$(document).ready(function() {
$('#contact_name').on('input', function() {
var input = $(this);
var is_name = input.val();
if (is_name) {
$('#contact_name').removeClass("has-success").addClass('has-error');
} else {
$('#contact_name').removeClass("has-error").addClass('has-success');
}
});
});
</script>
<h1>Example Form Validation</h1>
<br>
<div class="form " id="cla">
<?php $form=$this->beginWidget('CActiveForm',array( 'id' =>'form','htmlOptions'=> array('class'=> 'form-horizontal', 'name' => 'forvalidate', 'id' => 'registration-form' ))); ?>
<?php echo $form->errorSummary($model); ?>
<div class="row">
<div class="form-group col-lg-5" id='contact_name'>
<?php echo $form->label($model,'name',$htmloptions = array('class'=>'control-label ',)); ?>
<div class="controls">
<?php echo $form->textField($model,'name',array('name' =>'name' ,'class'=>'form-control col-xs-5', 'aria-describedby'=>'inputSuccess4Status')); ?>
</div>
</div>
</div>
<?php $this->endWidget(); ?>
</div>
<!-- form -->
Try
$('#contact_name').on('input', function() {
var input = $(this);
var is_name = input.val();
if (!is_name) {
$('#contact_name').parent().removeClass("has-success").addClass('has-error');
} else {
$('#contact_name').parent().removeClass("has-error").addClass('has-success');
}
});
I think "has-error" and "has-success" classes don't apply directly on the input but the div where the inputs are
You have to add these classes. See the highlighted code below:
$(document).ready(function() {
$('#contact_name').on('input', function() {
var is_name = $(this).val();
if (is_name) {
$('#contact_name').removeClass("has-success").addClass('has-error');
// ^^^^^^^^^^^^^^^^^^^^^^^
} else {
$('#contact_name').removeClass("has-error").addClass('has-success');
// ^^^^^^^^^^^^^^^^^^^^^^^^
}
});
});
I'm hoping of having this captcha on my website:
http://lirullu.com/
how would i test the outcome of this in a php form? Here is the code I have in my form.php file currently:
<script type="text/javascript">
$(window).load(function() {
var righthash1 = "cc03412089f87b9c509aaaa1";
$("#hf023d03").sortable({
items: ".cnt", stop: function () {
hash = "";
$("#hf023d03 .cnt").each(function (a, b) { hash += $(b).attr("id"); }); if (hash == righthash1) {
$(".slg1").fadeIn("fast");
} else {
$(".slg1").fadeOut("fast");
}
}
});
});
</script>
<div id="first_cnt">
<h2>Are you a traveline hero?</h2>
<span class="htext">(drag the answer)</span>
<div class="d1" id="hf023d03">
<div class="cnt" id="509aaaa1"><span class="yes">S</span></div>
<div class="cnt" id="89f87b9c"><span class="yes">E</span></div>
<div class="cnt" id="cc034120"><span class="yes">Y</span></div>
</div>
<div class="slg1">
<h3>is the correct answer!</h3>
</div>
</div>
I would seriously consider using Google's reCaptcha which has possibly the nicest API.
https://www.google.com/recaptcha
You can follow the tutorials and guides, it provides you the PHP and the Javascript you need to integrate the system with any website you create.
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.
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.