Develop Reference

Render tinymce 4 in a container display:none

I want to render tinymce4 on a textarea which is contained in a masked div but I have this error :
FontFamily is undefined
My render script is :
new tinymce.Editor('textarea_id', custom_settings, tinymce.EditorManager).render();
You can see the error in this fiddle, when you click on the "render" button :
http://fiddle.tinymce.com/uLfaab
With tinymce 3 I have no problem to do this.
Is there a way to solve this problem?
Thanks

If the <textarea> that contains TinyMCE is inside of a non-visible <div> then TinyMCE won't be visible either.
If you look at the source of the page (using browser tooling) you will see that TinyMCE is indeed "running" but not visible.
If you want the <textarea> to initially be invisible and then become visible when the button is clicked you can do that via JavaScript in your function behind your button.
I have updated your TMCE Fiddle: http://fiddle.tinymce.com/uLfaab/1
There were two changes...
1 - put an ID on your containing <div> so you can easily target it via JavaScript:
<form method="post" action="dump.php">
<div id="tmceDiv" style="display:none;">
<textarea id="tmce" name="content" class="mceEditorTxtModif"></textarea>
</div>
</form>
2 - Use JavaScript to make the containing <div> visible:
function test() {
new tinymce.Editor("tmce", tmce_settings_bureau, tinymce.EditorManager).render();
// This next line makes the DIV visible!
document.getElementById('tmceDiv').style.display = 'block';
}

contentEditable attr set to true makes links in content unclickable

I have a div with output content that sometimes outputs links.
But when this output div has the contentEditable attribute set to true if makes the links unclickable.
What's odd is you can still hover over it and before trying to click it it still looks and behaves like a link. (Blue in color, underlined, hover changes the color of blue, etc.)
But when you go to click it it doesn't take you to the link. I can right click and say open or open in a new window and it works fine.
Is there a way to make it clickable again so the user doesn't have to right click it all the time?
<div contentEditable='true'>
Some content with link: <a href='https://google.com' target='_blank'>Google</a>
</div>
Here is a JSFiddle example: https://jsfiddle.net/ce13w610/
I still need the content to remain editable, but it would be nice to have the links working intuitively again.
Edit: put additional solution as an answer, not an edit. :)

you have to make the link uneditable:
<div class='content-output' contentEditable='true'>
My content and link here: <a contentEditable='false' href='https://www.youtube.com/watch?v=dQw4w9WgXcQ' target='_blank'>Some video</a>
</div>
https://jsfiddle.net/ce13w610/1/

Here's an another one may be overkill though:
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div id="content" contentEditable="true">
Click Me!
</div>
<script>
$('#content a').mouseenter(function() {
console.log("Works");
$('#content').attr("contentEditable", "false");
});
$('#content a').mouseleave(function() {
$('#content').attr("contentEditable", "true");
});
</script>
https://jsfiddle.net/9dk4sgkk/

Additional solution:
I went and used jQuery to make all tags in the output div to have the attribute contentEditable = false.
$("#content-output a[href]").attr('contentEditable', 'false');

Image hide and show not working

I am using this jquery.smoothZoom.min.js to zoom and pan image.I have successfully applied that to my project for single image,now i want to add (<,>.i.e. corousal) ,so that I can use it for multiple images.When I add the corresponding part in my custom.js it does not work properly.
I will attach two screen sorts which will clear the picture
This is the first case
and after clicking the right corousal button
I can see only the background but not the required image . I can not understand what i am missing ,
This the html part i have been using
<div class="image-display" id="displayplan4" style="width:70%;height:120%; left:39%;top:10%;position:absolute;display:none;">
<img src="images/amenities.jpg" style="width:150%;height:130%; left:-60%;top:-20%;position:absolute;overflow:auto; z-index:1;">
<div style="width:150%;height:130%; left:-60%;top:-20%;position:absolute;background:rgba(255,255,255,0.7);z-index:2;">
</div>
<img class="planzoom" src="gallery/Residential/ongoing/Almog/Plan/almog1.jpg" id = "almogplan0" style="width:100%;height:100%; right:3%;top:50%;position:absolute;z-index:3;">
<!--button for forward and backward movement-->
</div>
and
<div id = "almogplandivII">
<img class="planzoom" src="gallery/Residential/ongoing/Almog/Plan/almog2.jpg" id= "almogplan1" style="width:100%;height:100%; right:3%;top:50%;position:absolute;z-index:3;display:none;">
</div>
and the corresponding js part to show and hide image on mouse click upon the image.
var almog_plan_div=0;
//Function for image forward with forward button
$("#almogforward").click(function ()
{
if(almog_plan_div<1)
{
$("#almogplan"+almog_plan_div).hide();
almog_plan_div++;
$("#almogplan"+almog_plan_div).show();
}
else
{
$("#almogplan"+almog_plan_div).hide();
almog_plan_div=0;
$("#almogplan"+almog_plan_div).show();
}
});
//Function for image backward with backward button
$("#almogback").click(function ()
{
if(almog_plan_div>0)
{
$("#almogplan"+almog_plan_div).hide();
almog_plan_div--;
$("#almogplan"+almog_plan_div).show();
}
else
{
$("#almogplan"+almog_plan_div).hide();
almog_plan_div=1;
$("#almogplan"+almog_plan_div).show();
}
});
I have tried like adding display:none style properties but it does not help my cause,
any help on this ?

Remove inline styling display: none from the img tag and then when u initialize your page, then hide that image using hide() function.
Inline-styling might be overriding this

Thanks to both of your answers ,both of u may not be exactly correct but definitely helped me getting me my solution.
The trick i was missing:
I) have used two different divs ,but i have not positioned the second one, (I noticed only that when I tried the whole thing in a new web page with only 2 images in it ,they were not positioned properly )my requirement needs the divs to be hidden ,i had to hide them.
2) The other thing i had to remove was the position:absolute thing from individual image elements.
Once rectified its cool now.

independently working div in Jquery

I am trying to make an independently working div which has a form inside of it.
I use jquery to calculate the price of a product depending of the user's selections in the form. However the user is able to add multiple items in his 'cart' so the form is duplicated to another div. The problem is that the calculation pattern can't separate these two divs and the calculation will be incorrect. The form is also interactive so it will be generated by the user's input. This is really complex set and renaming every variable by the 'product number' doesn't sound really efficient to me.
I'm kind of stuck here and i don't really know how to solve this problem. I had an idea that what if I put an iframe inside of the div and load my form and its calculation script inside of it, and then use post command to transfer the price of the product to the 'main page' to calculate the total price of all of the products the user wanted.
However it seems that jQuery scripts doesn't work independently inside of these iframes, they still have connection so they broke each other.
i will appreciate any kind of suggestions and help to solve this matter, thank you!
here's the code so far
Heres the body
var productNumber = 1;
<div id="div_structure">
</div>
<button id="newProduct" >Add new product</button><br \>
add new item
<!-- language: lang-javascript -->
$('#newProduct').click(function ()
{
$('<div id="productNo'+productNumber+'">')
.appendTo('#div_structure')
.html('<label onclick="$(\'#div_productNo'+productNumber+'\').slideToggle()">Product '+productNumber +' </label>'+
'<button onclick="$(\'#product'+productNumber+'\').remove()">Remove</button>');
$('<div id="div_product'+productNumber+'" style="display: none;">').appendTo('#product'+productNumber+'');
$('<iframe src="productform.html" seamless frameborder="0" crolling="no" height="600" width="1000">').appendTo('#div_product'+productNumber+'');
productNumber++;
});
it also has a function that allows the user to remove the inserted div.
Here's just few lines from the productform
$(document).ready(function()
{
$('#productCalculation').change(function ()
{
shape = $('input[name=productShape]:checked', '#productCalculation').val();
alert(shape);
});
});
<form id="productCalculation">
<div id="div_productShape" class="product1">
<h1>Select the shape of the product</h1>
<input type="radio" name="productShape" value="r1">R1</input><br \>
<input type="radio" name="productShape" value="r2">R2</input><br \>
<input type="radio" name="productShape" value="r3">R3</input><br \>
</div>
.
.
.
</form>
I translated all of the variables so they may not function correctly since i didn't test the translated version. So the problem is, if i try to make selections in the second generated div it wont even alert() the selected variable

There are two problems with this code: You say somewhere "I translated all of the variables so they may not function correctly since i didn't test the translated version. So the problem is, if i try to make selections in the second generated div it wont even alert() the selected variable". This is because event handlers are attached to elements that are in the DOM at that specific moment. To get it to work for all elements, use event delegation:
$(document).ready(function()
{
$(document).on( 'change', '#productCalculation', function ()
{
shape = $('input[name=productShape]:checked', '#productCalculation').val();
alert(shape);
});
});
Your other question is "My question in a nutshell: Is there a way to restrict jquery to function only in certain div even though i use the same variable names in the second div ". You can use the this variable to access the element the click was invoked on. From this element you can traverse the DOM if needed, for example with .parent().
$('div').on( 'change', function( e ) {
console.log( $(this).val() );
} );

How can accomplish a dual div swap on rollover?

I am looking for advice on a way to accomplish this. Given the following:
[Div A (contains an image)]
[Div B (contains a horizontal list of 8 or so text links)]
[Div C (contains text)]
Upon rolling over any link in Div B, how can I have Div A and Div C swap their respective contents out to something different that corresponds to the content of that link?
For example, if one were to rollover a Div B link called "Dogs", then upon that rollover, Div A would replace its contents and display an image of a dog and Div C would replace its contents and display text about dogs.
After rolling over that link, the new Div A and Div C contents will remain in place until a new link is rolled over.
I hope this makes sense. Does anyone have advice on the best way to accomplish this?

Assuming the href points to one resource that contains the content for both, but you can't just inject the entire output of the link into one element, something like this could work:
$('#divB a').mouseover(function() {
//get images from link, inject into divA
$('#divA').html('<strong>Loading...</strong>')
.load($(this).attr('href') + ' img');
//get divs from link, inject into divC
$('#divC').html('<strong>Loading...</strong>')
.load($(this).attr('href') + ' div');
});

Hmm... this should be pretty simple with jQuery (compared to some of the other answers here):
If you're unfamiliar with jQuery, the $() is a shortcut for calling jQuery(), and using
$(document).ready(function() {
// put all your jQuery goodness in here.
});
is a way to make sure jQuery fires at the right time. Read more about that here.
So first, add a class (ie .dogs) to each <a> element in your #divB list. Next, give each of the corresponding images the same class, and contain each of your text blocks in #divC in divs with the same class as well. The HTML would look something like this:
<div id="divA">
<img src="dogs.jpg" class="dogs" />
<img src="flowers.jpg" class="flowers" />
<img src="cars.jpg" class="cars" />
</div>
<div id="divB">
<ul>
<li>Dogs</li>
<li>Flowers</li>
<li>Cars</li>
</ul
</div>
<div id="divC">
<div class="dogs"><p>Text about dogs.</p></div>
<div class="flowers"><p>Text about flowers.</p></div>
<div class="cars"><p>Text about cars.</p></div>
</div>
Then use the following jQuery, putting this up in the <head> section of your HTML doc:
$(document).ready(function() {
$('a.dogs').hover(function() {
$('#divA img').hide("fast");
$('#divA img.dogs').show("fast");
$('#divC div').hide("fast");
$('div.dogs').show("fast");
});
});
We say when the document is ready, when you hover over the <a> element with the .dogs class, perform a function. That function will hide all of the images in #divA and immediately show the image with the .dogs class. Then it will hide all of the divs in the #divC and immediately show the div with the .dogs class.
You can do the same thing twice more for .flowers and .cars, or however many you have.
Keep in mind, there are more efficient ways of doing this too, if you're interested in looking deeper into jQuery, but this will be a solid way to get started in helping you understand exactly what jQuery is doing. And it keeps the script OUT of the HTML body, too!

You can change a div's contents with something like this:
<script type="text/javascript">
function over() {
var a = document.getElementById('a');
var c = document.getElementById('c');
a.style.backgroundImage = "url(/path/to/image)";
c.innerHTML = "<b>Dogs rock</b>";
}
</script>
<div id="a"></div>
<div id="b" onmouseover="over();"></div>
<div id="c"></div>
Then all you need to do is add whatever other div's you want and write code to change them appropriately. Set the initial state of A and C using css, or just call the over() function on page load.