complete

.idea/dataSources.xml

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+<?xml version="1.0" encoding="UTF-8"?>
+<project version="4">
+  <component name="DataSourceManagerImpl" format="xml" multifile-model="true">
+    <data-source source="LOCAL" name="@localhost" uuid="3a58a6ef-d5bc-4ae6-915f-796d28da868b">
+      <driver-ref>mysql</driver-ref>
+      <synchronize>true</synchronize>
+      <jdbc-driver>com.mysql.jdbc.Driver</jdbc-driver>
+      <jdbc-url>jdbc:mysql://localhost:3306</jdbc-url>
+      <driver-properties>
+        <property name="autoReconnect" value="true" />
+        <property name="zeroDateTimeBehavior" value="convertToNull" />
+        <property name="tinyInt1isBit" value="false" />
+        <property name="characterEncoding" value="utf8" />
+        <property name="characterSetResults" value="utf8" />
+        <property name="yearIsDateType" value="false" />
+      </driver-properties>
+    </data-source>
+  </component>
+</project>

.idea/sqldialects.xml

@@ -0,0 +1,9 @@
+<?xml version="1.0" encoding="UTF-8"?>
+<project version="4">
+  <component name="SqlDialectMappings">
+    <file url="file://$PROJECT_DIR$/pokemon_sql/pokemon_pokemon_trainer.sql" dialect="MySQL" />
+    <file url="file://$PROJECT_DIR$/pokemon_sql/pokemon_pokemons.sql" dialect="MySQL" />
+    <file url="file://$PROJECT_DIR$/pokemon_sql/pokemon_trainers.sql" dialect="MySQL" />
+    <file url="file://$PROJECT_DIR$/pokemon_sql/pokemon_types.sql" dialect="MySQL" />
+  </component>
+</project>

.idea/vcs.xml

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+<?xml version="1.0" encoding="UTF-8"?>
+<project version="4">
+  <component name="VcsDirectoryMappings">
+    <mapping directory="" vcs="Git" />
+  </component>
+</project>

Part1

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+I connected.

Part2

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+### Part 2: Simple Selects and Counts
+
+Directions: Write a sql query or sql queries that can answer the following questions
+
+* What are all the types of pokemon that a pokemon can have?
+
+        SELECT COUNT(id) FROM pokemon.types;
+
+* What is the name of the pokemon with id 45?
+
+        SELECT name FROM pokemon.pokemons WHERE id = 45;
+
+* How many pokemon are there?
+
+        SELECT COUNT(id) FROM pokemon.pokemons;
+
+* How many types are there?
+
+* How many pokemon have a secondary type?
+
+        SELECT COUNT(id) FROM pokemon.pokemons WHERE secondary_type IS NOT NULL;

Part3

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+Directions: Write a sql query or sql queries that can answer the following questions
+
+
+* What is each pokemon's primary type?
+
+        SELECT pokemon.pokemons.name, pokemon.types.name
+        FROM pokemon.pokemons
+        JOIN pokemon.types ON pokemons.primary_type = types.id;
+
+* What is Rufflet's secondary type?
+
+        SELECT pokemons.name, pokemons.secondary_type, types.name
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemons.secondary_type = types.id
+        WHERE pokemons.name = "Rufflet";
+
+* What are the names of the pokemon that belong to the trainer with trainerID 303?
+
+        SELECT pokemons.name FROM pokemon.pokemons
+        JOIN pokemon.pokemon_trainer
+        ON pokemons.id = pokemon_trainer.pokemon_id
+        WHERE pokemon_trainer.trainerID = 303;
+
+* How many pokemon have a secondary type `Poison`
+
+        SELECT pokemon.types.name, COUNT(pokemon.pokemons.name) AS "Count - Type Poison"
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemon.pokemons.secondary_type = pokemon.types.id
+        WHERE pokemon.types.name = "Poison";
+
+* What are all the primary types and how many pokemon have that type?
+
+        SELECT pokemon.types.name, COUNT(pokemon.pokemons.name) AS "Count"
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemons.primary_type = types.id
+        GROUP BY pokemon.types.id;
+
+* How many pokemon at level 100 does each trainer with at least one level 100 pokemone have? (Hint: your query should not display a trainer
+
+        SELECT COUNT(pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        WHERE pokemon.pokemon_trainer.pokelevel = 100;
+
+        ...Not sure how to answer the "each" part of the question without showing trainers. So below is an alternative response that shows "each" trainer by trainerID.
+
+        SELECT COUNT(pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        WHERE pokemon.pokemon_trainer.pokelevel = 100
+        GROUP BY pokemon_trainer.trainerID;
+
+* How many pokemon only belong to one trainer and no other?
+  * the answer is 0.
+
+        SELECT COUNT(pokemon.pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        HAVING COUNT(pokemon_trainer.trainerID) = 1;

Part4

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+Sort the data by finding out which trainer has the strongest pokemon so that this will act as a ranking of strongest to weakest trainer. You may interpret strongest in whatever way you want, but you will have to explain your decision.
+
+Explanation::
+I sorted by average pokemon level of each trainers. Then trainers with the same level, I sorted by who has the most pokemon. Following that, i sorted in alphabetical order.
+
+SELECT ANY_VALUE(pokemon.pokemons.name),
+       pokemon.trainers.trainername,
+       ANY_VALUE(pokemon.pokemon_trainer.pokelevel),
+       ANY_VALUE(type1.name),
+       ANY_VALUE(type2.name)
+FROM pokemon.pokemons
+       CROSS JOIN pokemon.pokemon_trainer ON pokemon.pokemons.id = pokemon.pokemon_trainer.pokemon_id
+       CROSS JOIN pokemon.trainers ON pokemon.trainers.trainerID = pokemon.pokemon_trainer.trainerID
+       LEFT JOIN pokemon.types AS type1 ON pokemon.pokemons.primary_type = type1.id
+       LEFT JOIN pokemon.types AS type2 ON pokemon.pokemons.secondary_type = type2.id
+GROUP BY pokemon.trainers.trainername
+ORDER BY AVG(pokemon.pokemon_trainer.pokelevel) DESC, COUNT(pokemon.trainers.trainername) DESC, pokemon.trainers.trainername

README.md

@@ -22,23 +22,83 @@ From here you should have all the pokemon data in your mysql schema. Feel free t
 Directions: Write a sql query or sql queries that can answer the following questions
 
 * What are all the types of pokemon that a pokemon can have?
+
+        SELECT COUNT(id) FROM pokemon.types;
+        
 * What is the name of the pokemon with id 45?
+        
+        SELECT name FROM pokemon.pokemons WHERE id = 45;
+        
 * How many pokemon are there?
+
+        SELECT COUNT(id) FROM pokemon.pokemons;
+        
 * How many types are there?
+
 * How many pokemon have a secondary type?
 
+        SELECT COUNT(id) FROM pokemon.pokemons WHERE secondary_type IS NOT NULL;
 ### Part 3: Joins and Groups
 Directions: Write a sql query or sql queries that can answer the following questions
 
 
 * What is each pokemon's primary type?
+
+        SELECT pokemon.pokemons.name, pokemon.types.name
+        FROM pokemon.pokemons
+        JOIN pokemon.types ON pokemons.primary_type = types.id;
+        
 * What is Rufflet's secondary type?
+
+        SELECT pokemons.name, pokemons.secondary_type, types.name
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemons.secondary_type = types.id
+        WHERE pokemons.name = "Rufflet";
+        
 * What are the names of the pokemon that belong to the trainer with trainerID 303?
+
+        SELECT pokemons.name FROM pokemon.pokemons
+        JOIN pokemon.pokemon_trainer
+        ON pokemons.id = pokemon_trainer.pokemon_id
+        WHERE pokemon_trainer.trainerID = 303;
+        
 * How many pokemon have a secondary type `Poison`
+
+        SELECT pokemon.types.name, COUNT(pokemon.pokemons.name) AS "Count - Type Poison"
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemon.pokemons.secondary_type = pokemon.types.id
+        WHERE pokemon.types.name = "Poison";
+                
 * What are all the primary types and how many pokemon have that type?
+
+        SELECT pokemon.types.name, COUNT(pokemon.pokemons.name) AS "Count"
+        FROM pokemon.pokemons
+        JOIN pokemon.types
+        ON pokemons.primary_type = types.id
+        GROUP BY pokemon.types.id;
+        
 * How many pokemon at level 100 does each trainer with at least one level 100 pokemone have? (Hint: your query should not display a trainer
-* How many pokemon only belong to one trainer and no other?
 
+        SELECT COUNT(pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        WHERE pokemon.pokemon_trainer.pokelevel = 100;
+        
+Not sure how to answer the "each" part of the question without showing trainers. So below is an alternative response that shows "each" trainer by trainerID.
+
+        SELECT COUNT(pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        WHERE pokemon.pokemon_trainer.pokelevel = 100
+        GROUP BY pokemon_trainer.trainerID;
+        
+* How many pokemon only belong to one trainer and no other?
+  * the answer is 0.
+ 
+        SELECT COUNT(pokemon.pokemon_trainer.pokemon_id)
+        FROM pokemon.pokemon_trainer
+        HAVING COUNT(pokemon_trainer.trainerID) = 1;
+        
 ### Part 4: Final Report
 
 Directions: Write a query that returns the following collumns: