Fix Typos

Author: WebDevSimplified

README.md

@@ -23,7 +23,6 @@ This table should be called `songs` and have four properties with these exact na
 After successfully creating the table copy the code from [data.sql](data.sql) into MySQL Workbench, and run it to populate all of the data for the rest of the exercises. If you do not encounter any errors, then your answer is most likely correct.
 
 ### 2. Select only the Names of all the Bands
-[Results](results/2.txt)  
 [Solution](solutions/2.sql)
 
 Change the name of the column the data returns to `Band Name`
@@ -100,19 +99,19 @@ If you performed this correctly you should be able to now see that band and albu
 ### 9. Delete the Band and Album you added in #8
 [Solution](solutions/9.sql)
 
-The order of how you delete the record is important since album has a foreign key to band.
+The order of how you delete the records is important since album has a foreign key to band.
 
 ### 10. Get the Average Length of all Songs
 [Solution](solutions/10.sql)
 
 Return the average length as `Average Song Duration`.
 
-| Average Song Length | 
-|---------------------| 
-| 5.352472513259112   | 
+| Average Song Duration | 
+|-----------------------| 
+| 5.352472513259112     | 
 
 
-### 11. Select the longest Song Length of each Album
+### 11. Select the longest Song off each Album
 [Solution](solutions/11.sql)
 
 Return the album name as `Album`, the album release year as `Release Year`, and the longest song length as `Duration`.
@@ -138,10 +137,10 @@ Return the album name as `Album`, the album release year as `Release Year`, and
 | Break the Silence           | 2011         | 6.15     | 
 | Tribe of Force              | 2010         | 8.38333  | 
 
-### 11. Get the number of Songs for each Band
+### 12. Get the number of Songs for each Band
 [Solution](solutions/12.sql)
 
-This is the toughest question on the list. It will require you to chain together two joins instead of just one.
+This is one of the toughest question on the list. It will require you to chain together two joins instead of just one.
 
 Return the band name as `Band`, the number of songs as `Number of Songs`.
 

schema.sql

@@ -2,13 +2,13 @@ CREATE DATABASE record_company;
 USE record_company;
 
 CREATE TABLE bands (
-	id INT NOT NULL AUTO_INCREMENT,
-	name VARCHAR(255) NOT NULL,
+  id INT NOT NULL AUTO_INCREMENT,
+  name VARCHAR(255) NOT NULL,
   PRIMARY KEY (id)
 );
 
 CREATE TABLE albums (
-	id INT NOT NULL AUTO_INCREMENT,
+  id INT NOT NULL AUTO_INCREMENT,
   name VARCHAR(255) NOT NULL,
   release_year INT,
   band_id INT NOT NULL,

solutions/1.sql

@@ -1,5 +1,5 @@
 CREATE TABLE songs (
-	id INT NOT NULL AUTO_INCREMENT,
+  id INT NOT NULL AUTO_INCREMENT,
   name VARCHAR(255) NOT NULL,
   length FLOAT NOT NULL,
   album_id INT NOT NULL,

solutions/10.sql

@@ -1,2 +1,2 @@
-SELECT AVG(length) as 'Average Song Length'
+SELECT AVG(length) as 'Average Song Duration'
 FROM songs;

solutions/11.sql

@@ -1,5 +1,5 @@
 SELECT
-	albums.name AS 'Album',
+  albums.name AS 'Album',
   albums.release_year AS 'Release Year',
   MAX(songs.length) AS 'Duration'
 FROM albums

solutions/12.sql

@@ -1,5 +1,5 @@
 SELECT
-	bands.name AS 'Band',
+  bands.name AS 'Band',
   COUNT(songs.id) AS 'Number of Songs'
 FROM bands
 JOIN albums ON bands.id = albums.band_id

solutions/4.sql

@@ -1,3 +1,14 @@
+
+/* This assummes all bands have a unique name */
 SELECT DISTINCT bands.name AS 'Band Name'
 FROM bands
-JOIN albums ON bands.id = albums.band_id;
+JOIN albums ON bands.id = albums.band_id;
+
+/* If bands do not have a unique name then use this query */
+/* 
+  SELECT bands.name AS 'Band Name'
+  FROM bands
+  JOIN albums ON bands.id = albums.band_id
+  GROUP BY albums.band_id
+  HAVING COUNT(albums.id) > 0;
+*/

solutions/5.sql

@@ -1,4 +1,4 @@
-SELECT DISTINCT bands.name AS 'Band Name'
+SELECT bands.name AS 'Band Name'
 FROM bands
 LEFT JOIN albums ON bands.id = albums.band_id
 GROUP BY albums.band_id

solutions/6.sql

@@ -1,5 +1,5 @@
 SELECT
-	albums.name as Name,
+  albums.name as Name,
   albums.release_year as 'Release Year',
   SUM(songs.length) as 'Duration'
 FROM albums

solutions/7.sql

@@ -1,5 +1,7 @@
 /* This is the query used to get the id */
-/* SELECT * FROM albums where release_year IS NULL; */
+/*
+  SELECT * FROM albums where release_year IS NULL;
+*/
 
 UPDATE albums
 SET release_year = 1986

solutions/8.sql

@@ -2,8 +2,10 @@ INSERT INTO bands (name)
 VALUES ('Favorite Band Name');
 
 /* This is the query used to get the band id of the band just added */
-/* SELECT id FROM bands
-ORDER BY id DESC LIMIT 1; */
+/*
+  SELECT id FROM bands
+  ORDER BY id DESC LIMIT 1;
+*/
 
 INSERT INTO albums (name, release_year, band_id)
 VALUES ('Favorite Album Name', 2000, 8);

solutions/9.sql

@@ -1,13 +1,17 @@
 /* This is the query used to get the album id of the album added in #8 */
-/* SELECT id FROM albums
-ORDER BY id DESC LIMIT 1; */
+/*
+  SELECT id FROM albums
+  ORDER BY id DESC LIMIT 1;
+*/
 
 DELETE FROM albums
 WHERE id = 19;
 
 /* This is the query used to get the band id of the band added in #8 */
-/* SELECT id FROM bands
-ORDER BY id DESC LIMIT 1; */
+/*
+  SELECT id FROM bands
+  ORDER BY id DESC LIMIT 1;
+*/
 
 DELETE FROM bands
 WHERE id = 8;