A path may contain the drive name, directory name(s) and the filename. To extract filename from the file, we use “GetFileName()” method of “Path” class. This method is used to get the file name and extension of the specified path string. The returned value is null if the file path is null.
To get current file's full path, you can use the os. path. abspath function. If you want only the directory path, you can call os.
You can retrieve the saved file path in the uploader success event and assign it to custom attribute (data-file-name) value of the respective file list element to open the uploaded file. Click the respective file element to create a new request along with saved file path using http header.
For security reasons browsers do not allow this, i.e. JavaScript in browser has no access to the File System, however using HTML5 File API, only Firefox provides a mozFullPath
property, but if you try to get the value it returns an empty string:
$('input[type=file]').change(function () {
console.log(this.files[0].mozFullPath);
});
https://jsfiddle.net/SCK5A/
So don't waste your time.
edit: If you need the file's path for reading a file you can use the FileReader API instead. Here is a related question on SO: Preview an image before it is uploaded.
Try This:
It'll give you a temporary path not the accurate path, you can use this script if you want to show selected images as in this jsfiddle example(Try it by selectng images as well as other files):-
JSFIDDLE
Here is the code :-
HTML:-
<input type="file" id="i_file" value="">
<input type="button" id="i_submit" value="Submit">
<br>
<img src="" width="200" style="display:none;" />
<br>
<div id="disp_tmp_path"></div>
JS:-
$('#i_file').change( function(event) {
var tmppath = URL.createObjectURL(event.target.files[0]);
$("img").fadeIn("fast").attr('src',URL.createObjectURL(event.target.files[0]));
$("#disp_tmp_path").html("Temporary Path(Copy it and try pasting it in browser address bar) --> <strong>["+tmppath+"]</strong>");
});
Its not exactly what you were looking for, but may be it can help you somewhere.
You cannot do so - the browser will not allow this because of security concerns.
When a file is selected by using the input type=file object, the value of the value property depends on the value of the "Include local directory path when uploading files to a server" security setting for the security zone used to display the Web page containing the input object.
The fully qualified filename of the selected file is returned only when this setting is enabled. When the setting is disabled, Internet Explorer 8 replaces the local drive and directory path with the string C:\fakepath\ in order to prevent inappropriate information disclosure.
And other
You missed );
this at the end of the change event function.
Also do not create function for change event instead just use it as below,
<script type="text/javascript">
$(function()
{
$('#fileUpload').on('change',function ()
{
var filePath = $(this).val();
console.log(filePath);
});
});
</script>
You can't. Security stops you for knowing anything about the filing system of the client computer - it may not even have one! It could be a MAC, a PC, a Tablet or an internet enabled fridge - you don't know, can't know and won't know. And letting you have the full path could give you some information about the client - particularly if it is a network drive for example.
In fact you can get it under particular conditions, but it requires an ActiveX control, and will not work in 99.99% of circumstances.
You can't use it to restore the file to the original location anyway (as you have absolutely no control over where downloads are stored, or even if they are stored) so in practice it is not a lot of use to you anyway.
Did you mean this?
$('#i_file').change( function(event) {
var tmppath = URL.createObjectURL(event.target.files[0]);
$("img").fadeIn("fast").attr('src',tmppath);
});
You can use the following code to get a working local URL for the uploaded file:
<script type="text/javascript">
var path = (window.URL || window.webkitURL).createObjectURL(file);
console.log('path', path);
</script>
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