In JavaScript, everything is 'truthy' or 'falsy', and for numbers 0
means false
, everything else true
. So you could write:
if ($(selector).length)
You don't need that >0
part.
With HTML5 you can make file uploads with Ajax and jQuery. Not only that, you can do file validations (name, size, and MIME type) or handle the progress event with the HTML5 progress tag (or a div). Recently I had to make a file uploader, but I didn't want to use Flash nor Iframes or plugins and after some research I came up with the solution.
The HTML:
<form enctype="multipart/form-data">
<input name="file" type="file" />
<input type="button" value="Upload" />
</form>
<progress></progress>
First, you can do some validation if you want. For example, in the .on('change')
event of the file:
$(':file').on('change', function () {
var file = this.files[0];
if (file.size > 1024) {
alert('max upload size is 1k');
}
// Also see .name, .type
});
Now the $.ajax()
submit with the button's click:
$(':button').on('click', function () {
$.ajax({
// Your server script to process the upload
url: 'upload.php',
type: 'POST',
// Form data
data: new FormData($('form')[0]),
// Tell jQuery not to process data or worry about content-type
// You *must* include these options!
cache: false,
contentType: false,
processData: false,
// Custom XMLHttpRequest
xhr: function () {
var myXhr = $.ajaxSettings.xhr();
if (myXhr.upload) {
// For handling the progress of the upload
myXhr.upload.addEventListener('progress', function (e) {
if (e.lengthComputable) {
$('progress').attr({
value: e.loaded,
max: e.total,
});
}
}, false);
}
return myXhr;
}
});
});
As you can see, with HTML5 (and some research) file uploading not only becomes possible but super easy. Try it with Google Chrome as some of the HTML5 components of the examples aren't available in every browser.
Best Answer
Basic usage of
.ajax
would look something like this:HTML:
jQuery:
Note: Since jQuery 1.8,
.success()
,.error()
and.complete()
are deprecated in favor of.done()
,.fail()
and.always()
.Note: Remember that the above snippet has to be done after DOM ready, so you should put it inside a
$(document).ready()
handler (or use the$()
shorthand).Tip: You can chain the callback handlers like this:
$.ajax().done().fail().always();
PHP (that is, form.php):
Note: Always sanitize posted data, to prevent injections and other malicious code.
You could also use the shorthand
.post
in place of.ajax
in the above JavaScript code:Note: The above JavaScript code is made to work with jQuery 1.8 and later, but it should work with previous versions down to jQuery 1.5.