Reputation: 11452
It must be very easy but I have no clue how to get the URL of an image from server.
Lets say I have domain named www.example.com and the image has been stored in my
htdocs>www>resources>userid>images>image.jpg
So, I want upload an image and save it through php script into my images folder and return the URL of that image back to browser like,
www.example.com/.../.../images/image.jpg
Is there any php function to get URL of a file?
After looking at answer I thought I should more clarify my self. I am really sorry for misunderstanding,
The help I wanted is to : what if the domain name isn't static I mean you can download and attach my image uploader plugin to your website but now it should dynamically get domain name and http path to your image.
Upvotes: 2
Views: 7944
Reputation: 6003
Since i don't have your plugin, i don't know the flow of your code. My guess below, hopefully, you can modify it according to your code.
Suppose i download your plugin to my website and upload an image using it. You can get the image path like this:
<?php
$upload_path = 'PATH_WHERE_IMAGE_IS_UPLOADED';
$image_name = 'NAME_OF_THE_UPLOADED_IMAGE';
$relative_path = $upload_path . DS . $image_name;
$path = str_replace( $_SERVER['DOCUMENT_ROOT'], $_SERVER['SERVER_NAME']
, $relative_path );
$path = 'http://' . $path; // this is where image is uploaded
?>
Note: This assumes that the upload directory is within the webroot.
Hopefully, this will work.
Upvotes: 5
Reputation: 5340
You can look at manual page of PHP function move_uploaded_file(). If I understand your problem, solution is there.
Upvotes: 0
Reputation: 1903
Editted
If you have for example:
$domain = $_SERVER['SERVER_NAME'];
$uploadPath = "/htdocs/www/resources/userid/images/image.jpg"
// replace "/htdocs/www", with an empty string
$subdir = str_replace("/htdocs/www", "", $uploadPath);
$url = $domain.$subdir;
Upvotes: 0
Reputation: 2259
Try this
$image='name of your image uploaded';
$imageurl='http://www.example.com/images'.$image;
echo '<input type=text name="image" value="'.$imageurl.'" disabled />;
Upvotes: 0