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ptyadanaptyadana
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add Instagram Clone Project
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‎Instagram Clone Project/14.Instagram Clone Database.sql

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‎Instagram Clone Project/15.Instagram Clone Database - Challenges.sql

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/*We want to reward our users who have been around the longest.
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Find the 5 oldest users.*/
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SELECT * FROM users
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ORDER BY created_at
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LIMIT 5;
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/*What day of the week do most users register on?
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We need to figure out when to schedule an ad campgain*/
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SELECT date_format(created_at,'%W') AS 'day of the week', COUNT(*) AS 'total registration'
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FROM users
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GROUP BY 1
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ORDER BY 2 DESC;
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/*version 2*/
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SELECT
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DAYNAME(created_at) AS day,
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COUNT(*) AS total
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FROM users
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GROUP BY day
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ORDER BY total DESC
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LIMIT 2;
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/*We want to target our inactive users with an email campaign.
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Find the users who have never posted a photo*/
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SELECT username
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FROM users
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LEFT JOIN photos ON users.id = photos.user_id
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WHERE photos.id IS NULL;
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/*We're running a new contest to see who can get the most likes on a single photo.
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WHO WON??!!*/
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SELECT users.username,photos.id,photos.image_url,COUNT(*) AS Total_Likes
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FROM likes
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JOIN photos ON photos.id = likes.photo_id
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JOIN users ON users.id = likes.user_id
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GROUP BY photos.id
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ORDER BY Total_Likes DESC
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LIMIT 1;
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/*version 2*/
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SELECT
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username,
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photos.id,
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photos.image_url,
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COUNT(*) AS total
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FROM photos
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INNER JOIN likes
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ON likes.photo_id = photos.id
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INNER JOIN users
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ON photos.user_id = users.id
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GROUP BY photos.id
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ORDER BY total DESC
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LIMIT 1;
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/*Our Investors want to know...
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How many times does the average user post?*/
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/*total number of photos/total number of users*/
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SELECT ROUND((SELECT COUNT(*)FROM photos)/(SELECT COUNT(*) FROM users),2);
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/*user ranking by postings higher to lower*/
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SELECT users.username,COUNT(photos.image_url)
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FROM users
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JOIN photos ON users.id = photos.user_id
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GROUP BY users.id
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ORDER BY 2 DESC;
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/*Total Posts by users (longer versionof SELECT COUNT(*)FROM photos) */
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SELECT SUM(user_posts.total_posts_per_user)
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FROM (SELECT users.username,COUNT(photos.image_url) AS total_posts_per_user
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FROM users
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JOIN photos ON users.id = photos.user_id
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GROUP BY users.id) AS user_posts;
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/*total numbers of users who have posted at least one time */
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SELECT COUNT(DISTINCT(users.id)) AS total_number_of_users_with_posts
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FROM users
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JOIN photos ON users.id = photos.user_id;
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/*A brand wants to know which hashtags to use in a post
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What are the top 5 most commonly used hashtags?*/
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SELECT tag_name, COUNT(tag_name) AS total
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FROM tags
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JOIN photo_tags ON tags.id = photo_tags.tag_id
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GROUP BY tags.id
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ORDER BY total DESC;
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/*We have a small problem with bots on our site...
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Find users who have liked every single photo on the site*/
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SELECT users.id,username, COUNT(users.id) As total_likes_by_user
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FROM users
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JOIN likes ON users.id = likes.user_id
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GROUP BY users.id
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HAVING total_likes_by_user = (SELECT COUNT(*) FROM photos);
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/*We also have a problem with celebrities
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Find users who have never commented on a photo*/
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SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NULL;
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SELECT COUNT(*) FROM
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(SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NULL) AS total_number_of_users_without_comments;
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/*Mega Challenges
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Are we overrun with bots and celebrity accounts?
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Find the percentage of our users who have either never commented on a photo or have commented on every photo*/
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SELECT tableA.total_A AS 'Number Of Users who never commented',
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(tableA.total_A/(SELECT COUNT(*) FROM users))*100 AS '%',
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tableB.total_B AS 'Number of Users who likes every photos',
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(tableB.total_B/(SELECT COUNT(*) FROM users))*100 AS '%'
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FROM
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(
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SELECT COUNT(*) AS total_A FROM
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(SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NULL) AS total_number_of_users_without_comments
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) AS tableA
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JOIN
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(
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SELECT COUNT(*) AS total_B FROM
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(SELECT users.id,username, COUNT(users.id) As total_likes_by_user
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FROM users
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JOIN likes ON users.id = likes.user_id
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GROUP BY users.id
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HAVING total_likes_by_user = (SELECT COUNT(*) FROM photos)) AS total_number_users_likes_every_photos
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)AS tableB;
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/*Find users who have ever commented on a photo*/
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SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NOT NULL;
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SELECT COUNT(*) FROM
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(SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NOT NULL) AS total_number_users_with_comments;
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/*Are we overrun with bots and celebrity accounts?
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Find the percentage of our users who have either never commented on a photo or have commented on photos before*/
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SELECT tableA.total_A AS 'Number Of Users who never commented',
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(tableA.total_A/(SELECT COUNT(*) FROM users))*100 AS '%',
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tableB.total_B AS 'Number of Users who commented on photos',
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(tableB.total_B/(SELECT COUNT(*) FROM users))*100 AS '%'
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FROM
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(
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SELECT COUNT(*) AS total_A FROM
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(SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NULL) AS total_number_of_users_without_comments
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) AS tableA
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JOIN
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(
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SELECT COUNT(*) AS total_B FROM
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(SELECT username,comment_text
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FROM users
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LEFT JOIN comments ON users.id = comments.user_id
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GROUP BY users.id
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HAVING comment_text IS NOT NULL) AS total_number_users_with_comments
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)AS tableB

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