I am trying to target a specific div with the results of a form post.
I have found the below code, but am unclear on where the URL for the page that handles the form data is specified. Any help greatly appreciated.
$("form1").submit(function() {
$.post($(this).attr("action"), $(this).serialize(), function(html) {
$("#someDiv").html(html);
});
return false; // prevent normal submit
});
For example, I want to send form id="form1" to somepage.php, and have somepage.php displayed into div id=someDiv.
In that code-snippet, $(this).attr("action") is the URL: it's taken from the action="..." attribute of the <form> element. If, for whatever reason, you don't want to use that attribute to specify the URL, you can replace $(this).attr("action") with an explicit URL (as a string).
It is defined in the action attribute of your form tag
Related
Is it possible to use ckeditor on a form input instead of textarea, i am building a CMS and now trying to add ckeditor and majority of of fields are form input not textarea
Thanks in advance
You can use CKEditor on an element (a div, say) that has contenteditable set. In fact, by default contenteditable elements will have CKEditor editors instantiated. It seems unconventional to use a rich text editor on an input of type text but I imagine it could be done.
As far as I know CkEditor has to be created using a textarea. In saying this, I am using it in a Razor MVC view and its one of the classes in my form..
The request will be blocked though and you will get this error;
A potentially dangerous Request.Form value was detected from the
client
To get around this, you need to get the value of the CKEditor text and html encode it. You can do this
when submitting your form, intercept it on the onsubmit function:
<form id="myform" onsubmit="javascript:onFormSubmit(this);return false" >
...
</form>
Then in this onFormSubmit function
Get the value,
Set the property value to the url encoded value
Do a ajax call to your server with the data
Your function will get the CKEditor encoded value like so:
function onFormSubmit(form)
{
var editor = CKEDITOR.instances["EditorId"];
var richtextValue = editor.getData();
var urlEncodedValue = encodeURIComponent(richtextValue);
// TODO rest of code doing ajax post to submit your form and its data
// Here you need to do an ajax call to your server pass it the form data along with the url encoded CKEditor value for that property
}
My questions is - how do I add a class or id to the body tag using a text field within Orchard?
So if I enter the word "product" in the text field then the result should be <body class="product">. I want to use this method instead of creating alternate layout templates as every page has the same layout but I need a different class for each page to reference a different colour scheme I have setup for each page in my CSS.
I have added a text field with the name Area to the Title ContentType in the backend. My problem is now how to get the value of the field to be put into the body in the Document.cshtml.
I have read Setting Unique Body Classes and IDs in Orchard and Using Alternatives for Document.cshtml in Orchard CMS but I still can't get it to work! The second one seems like what I want to do but so far I have been unable to acheive it.
Any answer would be very appreciated?
Thanks
Andy
I am so frustrated that there is no readily available solution found in the net about this. Since I am more comfortable coding in jquery & js...
Here's my solution: Assuming you have jquery loaded...
#using(Script.Foot()) {
<script type ="text/javascript">
//<![CDATA[
$(document).ready(function () {
var url = window.location.pathname;
var count = url.match(new RegExp("/", 'g'));
var urlsplit = url.split("/");
var page_class = urlsplit[count.length];
alert(page_class);
$('body').addClass(page_class);
});
//]]>
</script>
}
The easiest way to achieve this is to use the Classy feature from Vandelay.Industries.
#UiField Hyperlink historyLink;
this.historyLink.getElement().setAttribute("rel", "nofollow");
This sets the rel attribute of the containing div.
<div class="gwt-Hyperlink" rel="nofollow">history</div>
How do I get at the a tag?
You can get a element this way
Element a = historyLink.getElement().getFirstChildElement();
Maybe you can use an InlineHyperlink instead (if you don't want the div wrapper at all), but actually I don't understand the use of rel=nofollow on a link with a #hash-only href: if that's for Google AJAX crawling, can't you simply avoid outputting the link in the HTML snapshot you send in response to an _escaped_fragment_ request?
I'm using jQuery Form (http://jquery.malsup.com/form/) to get send data to a form - is there a way I can then, without refresh, put the results page that's generated by the form into a div on the page?
Any advice appreciated!
I would suggest not using that form plugin - it was made back in the days before there was an easy way to serialize form data using jQuery, and serves no real purpose any more. I would suggest something like this:
$("form").submit(function() {
$.post($(this).attr("action"), $(this).serialize(), function(data) {
$("#someDiv").html(data);
});
return false; // prevent normal submit
});
If you insist on using jQuery Form Plugin - which is NOT recommended -, you can set the target option to a selector of the element(s) you would like to fill up:
// prepare Options Object
var options = {
target: '#divToUpdate',
url: 'comment.php',
success: function(data) {
alert('Thanks for your comment!');
}
};
Take a look at http://jquery.malsup.com/form/#options-object for more info.
To prevent refresh, just make sure the form's onsubmit event returns false:
<form method="post" onsubmit="return false">
I wanted to find a way to upload a single file*, in the background, have it start automatically after file selection, and not require a flash uploader, so I am trying to use two great mechanisms (jQuery.Form and JQuery MultiFile) together. I haven't succeeded, but I'm pretty sure it's because I'm missing something fundamental.
Just using MultiFile, I define the form as follows...
<form id="photoForm" action="image.php" method="post" enctype="multipart/form-data">
The file input button is defined as...
<input id="photoButton" "name="sourceFile" class="photoButton max-1 accept-jpg" type="file">
And the Javascript is...
$('#photoButton').MultiFile({
afterFileSelect: function(){
document.getElementById("photoForm").submit();
}
});
This works perfectly. As soon as the user selects a single file, MultiFile submits the form to the server.
If instead of using MultiFile, as shown above, let's say I include a Submit button along with the JQuery Form plugin defined as follows...
var options = {
success: respondToUpload
};
$('#photoForm').ajaxForm(options);
... this also works perfectly. When the Submit button is clicked, the form is uploaded in the background.
What I don't know how to do is get these two to work together. If I use Javascript to submit the form (as shown in the MultiFile example above), the form is submitted but the JQuery.Form function is not called, so the form does not get submitted in the background.
I thought that maybe I needed to change the form registration as follows...
$('#photoForm').submit(function() {
$('#photoForm').ajaxForm(options);
});
...but that didn't solve the problem. The same is true when I tried .ajaxSubmit instead of .ajaxForm.
What am I missing?
BTW: I know it might sound strange to use MultiFile for single-file uploads, but the idea is that the number of files will be dynamic based on the user's account. So, I'm starting with one but the number changes depending on conditions.
The answer turns out to be embarrassingly simple.
Instead of programmatically submitting using...
document.getElementById("photoForm").submit();
... I used...
$("#photoForm").submit();
Also, since I only need to upload multiple files on occasion, I used a simpler technique...
1) The form is the same as my original...
<form id="photoForm" action="image.php" method="post" enctype="multipart/form-data">
2) The file input field is basically the same...
<input id="photoFile" "name="sourceFile" style="cursor:pointer;" type="file">
3) If the file input field changes, submit is executed...
$("#photoFile").change(function() {
$("#photoForm").submit();
});
4) The AjaxForm listener does its thing...
var options = {
success: respondToUpload
};
$('#photoForm').ajaxForm(options);