I am getting the image links where my image is there by using reverse image search provided by google but now we want the exact path of that page where my particular image is situated like http://www.indiamart.com/lelijiye-bengaluru/maroon-color-desginer-saree.html page
Direct linking is the use of a particular image or multimedia file on another website by linking directly to it on another web host's server. (I use Photobucket)
For example, if you want to post an image on another forum, a direct linking URL (found on the edit image page) can be used to display that image via the PBase server on another website or forum.
An easy way to find the direct linking URL for your images is to go to the edit image page and see the 'direct linking URL' at the top right of the screen.
Related
New :
I have a wallpaper website and I recently checked the server log of the files accessed and I saw that I am getting lots of traffic on the image files rather than the page of the website.
So I want to do is redirect example.com/sample-image.jpg to example.com/sample-image.html
I am using WordPress CMS and I tried one .htaccess code but it redirects all the images to the homepage of website :(
OLD:
When a user attempts to load an image directly into the browser (for example from Google Images), the browser normally opens just the bare image.
I'm trying to address the following situation:
1) User is browsing Google Images and finds an image. 2) Google Images offers the user to open the website where the image was found or just open the original image. 3) The user chooses the latter, therefore the ball is now on the webserver where the image is hosted. 4) htaccess detects a human operated browser trying to access directly to an image in /uploads/ folder. 5) htaccess redirects the user to single-attachment.php or post in which image include or attached.
Images/media it is post type. Try get_permalink($id_of_attachment); If you have url of image you can get id too. Example by get_post_thumbnail_id();
Need to upload files to dropbox and get the public url of that file. I'am generating thumbnails and need to show that in my site. I should also show the original files to users without login to the dropbox site.
After uploading to dropbox I need to get the original file url which will be used by crocodoc api viewer.
I could able to upload the file and I am getting the metadata of that file. I created shared url and given that url that url img tag in html but seems returning as html instead of image.
I have gone through similar questions mentioned in SO but almost all are referring old documents.
I found this trick earlier and thought I'd share.
Go to: https://www.dropbox.com/home
Then, highlight your mouse over an item, click on share.
This will popup:
Simply change:
https://www.dropbox.com/s/io9oz94lgm2j2bb/loadingscreen.jpg?dl=0
to
https://dl.dropbox.com/s/io9oz94lgm2j2bb/loadingscreen.jpg
Replacing the www with dl, to get the direct, public url.
Not sure if this would fall here or Webmasters so I apologize in advance.
Currently I have an image viewer written in PHP that uses a rewrite to change:
http://example.com/image/s8d8s.png to http://example.com/s8d8s
When they load this viewer it will display the image with all kinds of links and stats for the image. It also connects to my MySQL database and adds a view to the stats table. So that I can show how many views each image has.
The issue: When a use right-clicks on an image in the viewer and gets the image's direct link(http://example.com/image/s8d8s.png) and uses that by sending it to a friend or embedding it on a website, I want to be able to track that and add a view to my database.
Is there a way I can manipulate the apache rewrite to trick a browser into thinking it's viewing a image direct link but it be a PHP file that will add a view?
I have a script where by users upload an image on a website and the image gets uploaded to an FTPServer. The name of the file is stored in a database, so when users click on a link, the query string is used as a reference in the Database to get the image name. The only part I'm stuck on right now, is how to display the image on my webpage using php. Is there a specific function to get a copy from the FTP server and display the image? I dont want do download the image to the web server, and then display the image and then delete it, as this could take up a lot of space if the web site has many visitors.
The "big image hosts" typically have web access to their "storage servers" as well. So when a file is uploaded to their image server it's accessible with a URL. You'd want to determine the url that points to the file you just uploaded and place that in your html (in the img tag. Let the user's browser retrieve the image, as opposed to your web server first retrieving it and then displaying the page.
You have to store the image on the web server, once you do that you can display it on your web page however you like, php, ajax, javascript, your choice.
I had trouble finding an answer for this problem. I would like to display a link to a page that has a video file on it but I don't want the link to appear until after the video file is uploaded.
It's a site where users upload videos so I don't want to have to manually put the links there for them after they upload their videos, nor do I want to make the links ahead of time since users will click them and there will be no video available.
So, is there a way to keep a link hidden until a file is uploaded. Once the file is uploaded the php page where the link will be on checks to see if it exists and then if it does the link is shown, otherwise the link stays hidden.
Each time the form that the client uses to upload a file is sent store the information of the file in a database.
Each time the client request the list of videos, query the database and build the web page based on the results.
Make sure to associate the uploaded files to the users, so you don't mix the up. Are you managing authentication already?
If you want to have the links appear as soon as the upload is finished without reloading the page, then you need ajax.