YouTube_Shorts_Video_Editing

Youtube shorts quizzes video editing using Web and Java backend

Using a front end of HTML CSS and Javascript Database of SQL and Backend of Java

Ill make a video editing program to automate video editing of my youtube videos.

Frontend Public - this is where all of the files that dont need to be compiled or processed go, like html and images

Src - this is where all my code and styling files go that need to be compiled and processed first.

Video Editing Audio - Xuggle (based on FFmpeg) or JCodec Video - Text - AWT and Swing

Use BufferedImage and ImageIO to store images and render each out individually into png frames

How to use the MariaDB Server

'''connect to the mariadb''' ~sudo mariadb -u root

'''Exit mariadb (run while mariadb is running)''' ~EXIT;

Here on out these are sql commands once mariadb is running

'''Show the databases''' ~SHOW DATABASES;

'''Select/use a database''' ~USE {database name};

'''Create a database''' ~CREATE DATABASE {database name};

'''Delete a database''' ~DROP DATABASE {database name};

'''Create a table''' ~CREATE TABLE {table name} ( id INT AUTO_INCREMENT PRIMARY KEY, name VARCHAR(100), age INT );

'''Insert data into a table''' ~INSERT INTO {table name} ({element1}, {element2}, ...) VALUES ({element1 value}, {element2 value}, ...); ~INSERT INTO example (name, age) VALUES ('John Doe', 30); ~INSERT INTO example (name, age) VALUES ('Jane Smith', 25);

'''Get data from a table''' selects the whole table and everything in it ~SELECT * FROM {table name};

selects just the name and age columns ~SELECT name, age FROM example;

this will grab all rows which have an age of greater than 10 ~SELECT * FROM example WHERE age > 10;

this will grab all names where the age is greater than 10 ~SELECT name FROM example WHERE age > 10;

this will grab all id and name where age is greater than 10 ~SELECT id, name FROM example WHERE age > 10;

you can chain ifs together ~SELECT FROM example WHERE age > 10 AND name < 100; ~SELECT FROM example WHERE age > 10 OR name = 'John Doe;

you can use a range statement ~SELECT * FROM example WHERE age BETWEEN 25 AND 35;

only grab rows that contain these elements ~SELECT * FROM example WHERE name IN ('John Doe', 'Jane Smith');

use the like keyword to not exactly search % will match an unlimited amount of characters _ will only match for one character ~SELECT FROM example WHERE name LIKE 'J%'; ~SELECT FROM example WHERE name LIKE '_ane Smith';

order the results in some way SELECT * FROM example WHERE age > 25 ORDER BY age DESC;

in this case there is 2 sorting criteria. so if there is 2 of the same name it will then sort by age SELECT name, age FROM example ORDER BY name ASC, age DESC;

math operations on the tables returns the count of all of the rows SELECT COUNT(*) FROM example;

from the rows returned it will average the age SELECT AVG(age) FROM example;

from the rows returned it will return the max age SELECT MAX(age) FROM example WHERE name LIKE 'J%';