postpic | open source dbms PostgreSQL that enables image processing | SQL Database library
kandi X-RAY | postpic Summary
kandi X-RAY | postpic Summary
PostPic is an extension for the open source PostgreSQL dbms that enables image processing inside the database, like PostGIS does for spatial data. It adds the new image type to the SQL, and several functions to process images and to extract their attributes. Traditional relational clauses can be combined in queries with image-related ones, and basic image processing can be performed in SQL.
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QUESTION
I make a site with multiple users, making posts with images and ability to add/remove friends. So it's easy to make two different pages for post list and creating a new one. But of course it looks better when you can read posts and make new at the same place.
As I understand (learn django for less than a month), I can't connect 2 views to the same url, so the most logical way I see is to join 2 views in one, I also tried to play with template inheriting to render post form by including template, but actually it doesn't work.
Below you can see my views, Post model, and templates. Thank you for attention.
views.py:
...ANSWER
Answered 2020-Oct-26 at 22:03You can put the post_create_form on the same page as post_list_view there is no need to make a separate view for post creation but You need to make ones for editing and deleting.
You can give all of these views the same HTML page with different URLs.
Using template_name = 'example/example.html' ,in Class_Based_Views.
I hope I understand your problem if not clarify more why you can't join two views in one.
QUESTION
I have a page, searchpage.php, that generates 'posts' dynamically for a user's search inputs.
...ANSWER
Answered 2020-Oct-23 at 04:08Use sessions. And enable it using:
QUESTION
I tried to upload a file(image) on firebase storage. but it consoles an error saying "Uncaught Error: No Storage Bucket defined in Firebase Options." . This is my code
...ANSWER
Answered 2017-Aug-07 at 14:51It's likely the "storageBucket" option in the Firebase config is an empty string. It seems to default to an empty string unless "Storage" has been initialized on the console interface.
Go to your firebase console, initialize "Storage" by accessing it on the Firebase console interface, and then copy the new value for "storageBucket" from the firebase config into your app.
Your firebase config should look something like this.
QUESTION
I'm trying to allow users to see the post of users they are following. The problem is that only the logged in users post are showing on the feed. My following node on firebase looks like:
Following Collection
...ANSWER
Answered 2019-Jul-14 at 16:28You may need to restructure your Posts node to something like this:
Posts
QUESTION
I am attempting to display my image's information dynamically within the data-caption attribute for fancybox3. It is quite long and I am unsure how change it so its not hideous. The specific problem is that when it grabs the information from the mysql database that may contain special characters, like ' or " the data-caption thinks it is being closed off when that is not what I would like.
I have tried separating the very long echo statement into multiple ones but that did not change anything. I tried to append strings for data-caption as such, " .$var ." and that has not worked. I have tried using mysqli_real_escape() during variable declarations and it does not change the functionality when trying to output the database information.
...ANSWER
Answered 2019-Apr-07 at 07:24htmlspecialchars() is the answer https://www.php.net/manual/en/function.htmlspecialchars.php
Also, learn to debug your code - if you would have used developer tools to check generated html code, I can guarantee that you would have seen that your html is broken (due to unescaped characters)
QUESTION
I got this form for adding new posts and i can't get it to write to DB. I debugged it as much as i can and when i try to add i don't get any errors but also i don't see the post stored in my DB. Am i missing something? Code:
-This is the form
...ANSWER
Answered 2018-Sep-16 at 11:50Your issue I suspect is you are checking if your image file has been uploaded with $_POST['image']. This isn't how PHP handles file uploads - they are stored in $_FILES instead so your isset($_POST['image'] == false.
This should work:
QUESTION
Good afternoon everyone, I am trying to upload an image to firebase and display back on a cell..I have just about all the content working except the image (i cant get rid of the error).
Here is the error I am getting
...Cannot invoke initializer for type 'Posts' with an argument list of type '(postImageStringUrl: String, content: String!, postId: String)'
ANSWER
Answered 2017-Oct-28 at 13:37Your Posts init is
QUESTION
[I have structured my database as below][1]
...ANSWER
Answered 2017-Jun-24 at 06:32Inside onDAtaChange,try like this ways and let me know
QUESTION
I am a student and I am trying to start, pause and start a gif when a user clicks the gif, however I am stuck on how to add in this click function. I know that the the gif version of the object is .images.fixed_height.url and the still image is .images.fixed_height_still.url . If I try to append like below $(this) I get that images is undefined. How would I go by doing this? Currently 10 gifs show when you click the category. Thank you for any help in advance.
Code:
...ANSWER
Answered 2017-Jun-01 at 04:50you can use this jquery plugin http://rubentd.com/gifplayer/
QUESTION
I'm currently reading images from my firebase storage - which works fine.
I have set up a caching to read images from the cache when it has been read from the storage:
...ANSWER
Answered 2017-Apr-27 at 11:54You can realize caching images with Kingfisher for example. And works better. link
How to use: Add link to your image from storage to database item node. Like this:
Then just use it to present and cache image.
Example:
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