so I started using tinymce for my blog posts and there is this problem with imagetools, they work while images are being edited, but after the article is posted, only this BIG url is showing. I think the problem is that, the blob image which is created while editing is not saved and I need to save it, right?
By the way I am using laravel.
When you edit images using the image tools they will always be turned into a Base64 or BLOB encoded image - the image tools function in the browser and know nothing (directly) of the application in use (e.g. Laravel). As such its up to the application to determine what to do with these images:
https://www.tinymce.com/docs/advanced/handle-async-image-uploads/
The basic process is that TinyMCE will create a separate HTTP POST for each image that you modify with the image editor. It will send that image to a URL of your choosing (via HTTP POST) based on the setting of the images_upload_url option in your init.
The image handler at the URL referenced in the images_upload_url (which you have to create) has to do whatever needs to be done to "store" the image in your application. That could mean something like:
Store the item in a folder on your web server
Store the item in a database
Store the item in an asset management system
I would assume Laravel has some APIs such as the Filesystem API that you could use to store the images appropriately.
Regardless of where you choose to store the image your image handler needs to return a single line of JSON telling TinyMCE the new location of the image. As referenced in the TinyMCE documentation this might look like:
{ location : '/uploaded/image/path/image.png' }
TinyMCE will then update the image's src attribute to the value you return. If you use the images_upload_base_path setting in the init that will be prepended to the returned location.
The net here is that TinyMCE knows when an embedded image exists in your content but it can't possibly know what to do with that image in the context of your application so that job (the "image handler") is something you must create.
Related
My question is about HTML and PHP.
This is my setup right now:
A website where user have accounts
A FTP server with pictures (currently none)
Files are currently saved on the website in the "PICTURES" folder (which is accessible by everybody who know the full URL)
So, I would like to know how I can display the images without storing them on the website (which will fix my URL problem).
My idea was to move the files on the FTP server, and when a users logon and request a page with those images, download them through a FTP connection, save them on the website, display the images, and remove them. Which would make them accessible only between the downloading time. But this solutions sounds REALLY bad to me.
You need always to have a place where your images are stored. But, if you don't want to give a user the chance to know where are stored, you can create a system which is used to show the images.
Think about this, if you want to download a file from Mega, you can't access to the URL where the file is stored, instead of that, the server itselfs calls a system who assign you a "key" and you can download the file only through that system using your "key".
You could use a system like "base64" so you can encode your image, and show it using it, or, you can use the "header" modifier so, you can display an image using a PHP code.
For example your image tag will be like:
<img src="processImage.php?id=01&user=10&key=123" />
So, your processImage will return a "tricky" image, actually not the image, but the code processed by PHP will be returned, like using "imagejpg()" function with the header "Content-Type:image/jpeg" and then the user will not know where the image is stored actually but the img will works actually.
I have an app that download some images from a server. Now app create a http post request for php script, that retrieve image (4 for each request) and sent it to my app into a json response (encoded in base 64). I put images into a ScrollView, and when user reach the end of the list, a new group of images is downloaded.
That is not the most perfotmant way, so i would use some LazyLoad libraries found on gitHub, but all of that require link to image, but i wouldn't sent in any way image link to app.
So, how can i do to retrieve images with lazy load?
Requesting images one by one or four by four does not make such a difference. In both cases you would first put a placeholder image in the imageview(s) and upon download complete and extracting done put the images in the respective imageviews if they do still exist. I suppose you put the downloaded and extracted images on the device so you can use them again when the user scrolls or starts the app again?
I'm using regular imagefields on various content types in a Drupal 7 social website (and also on the user profile pictures via http://drupal.org/project/imagecache_profiles). For cropping I'm using http://drupal.org/project/imagecrop as an image style effect.
I would like to replace the original image with the one generated from the imagecrop style. How do you replace the original image from a user (the one that gets used for subsequent image styles when you say call it in views) with one that is created from an image style effect?
I'm not sure the best way to approach this problem, which is why I posted question.
To make sure I'm being clear -
A User uploads a large image (say 2,000 pixels by 1300 pixels) via a regular imagefield.
Using image styles, I perform some effects (specifically javascript crop, but this could be generic) and then replace the original image with the one generated from the image style.
All additional image styles (different sizes etc) use the new original images and are derived from it.
I think this can be done easily using a custom module and they must have used something like the image editor project - See drupal.org/project/imageeditor, just not sure what hooks to use or how to jump in on this. Any advice, tips or direction would be great.
Check module - Original image with style
I believe you can address your problem by making changes under the display settings for your field:
/admin/structure/types/manage/[your content type machine name here]/display
Image this scenario:
There is a picture locally in my server, where a sketch is displayed, and there is a "blank hole" area on it.
Then, a user can upload another picture to my server.
What i'm trying to achieve is this:
After image upload is finished, the first image (the one with the "hole") is displayed, and behind it is displayed the user's photo, so that you can see it through the "blank hole" area of the first photo.
Then the user can move his picture (drag & drop style) so he can choose which area of it is visible through the "blank hole".
Then i would like to save the result - by merging the 2 photos or keeping the position of the user's picture in a db so i can display it again later.
(Something like this more or less)
What kind of technollogy should i look for? I'd guess javascript(for the drag & drop) or html5 or php(for merging the photo)?
Are there any libraries that i can use?
I hope my explanation isn't too messy, i didn't even know how to google for it.
I don't know if there are better solutions (and I suspect there are), but I suspect all of this can be done with not too much trouble. Here's a rundown of one way to approach the problem:
Use a JavaScript-powered "upload widget" such as uploadify to enable your user to upload "his" image to the server. The server will do some processing on the image (e.g. resize and crop to suitable dimensions) and save it using e.g. PHP's gd library. It will return a URL to the "prepared" image back to the browser -- all of this through AJAX.
The browser then has a URL to the user's image, so using more Javascript you can dynamically add an element that displays it inside the page and allow the user to move it around with e.g. jQuery draggable. Compositing the draggable image behind your static content (the image with the "hole") is a detail you will have to take care of using a combination of HTML, CSS and again Javascript.
When the user is done, use an AJAX call (e.g. again jQuery) to inform the server of the image's positioning (this will be available through the facilities of the Javascipt framework you have selected). The server can then "compose" the two images together (gd or something equivalent once more) and return to the browser a URL through which the final product can be accessed.
Of course there are lots of details to take care of here, but knowing exactly what the plan is should help you get started.
Have a look at the PHP GD extension. If it's installed, it's pretty easy to have an image (with a transparent center) to be merged on top of a second image that a user would upload.
Have a look at http://php.net/manual/en/function.imagecopymerge.php
Ok to get you started, yes use a JavaScript drag and drop module for the placing of the image. You can record the x /y cordinates relative to the container. Do the image merging with a PHP image library / Class. Something like this : http://www.phpclasses.org/package/3930-PHP-Generate-an-image-from-the-combination-of-2-images.html
I am creating a multi-layered image editor using AJAX to send calls to a PHP script, then using a GD library to process the changes, save the new image and send the new image path back to the browser for AJAX to replace old with new.
This example, when you move a layer it seamlessly moves the layer without a single flicker of the image. This example uses a dll file to serve the image, I want to be able to achieve the same thing in PHP / JS.
Idea 1
I thought about applying a z-index to the current image and load the new image below it, once it's loaded, hide the old image.
Idea 2
Set the image path to a php script which outputs the raw data of the compiled image direct to the browser.
if you want to replace the image "without a single flicker", just preload the image
(load the image to a hidden img element or a Image javascript object and then show it on an onload callback)