1. Creating Databases SELECT. UPDATE. Demonstrate projects. Do not track.
Classwork / Homework: Prepare to choose teams & projects.

2. Warning Changes in php from last time.
THIS IS INEVITABLE and is mainly a good thing.
I think I made required changes, but only did modest testing.
AND I did all the testing. Always involve other people to do testing!

3. SELECT … what if you want only DISTINCT values?
For example, the trivia quiz starts with a form to choose the category. People (site administrators) adding questions can put in any category.

4. from php code choosecategory.php
<h1> Welcome to the Quiz </h1> <br>
<h3> Sign in and select a category for your question </h3>
<form action="askquestion.php" method=post>
<p>Name <input type=text name='player' size=30
<?php …
$query="SELECT DISTINCT category FROM questions";
$categories = mysqli_query($link,$query);
while ($row=mysqli_fetch_array($categories))
{ $cat=$row['category'];
print ("<option value='$cat'>$cat</option><br>\n"); }

5. SQL join
SELECT statement against a table made up by JOINing tables together on identical fields.
Different types of JOIN:
JOIN (same as INNER JOIN)
LEFT JOIN
RIGHT JOIN
FULL JOIN 5

6. LEFT, RIGHT, FULL JOINs These provide ways to pick up missing records.
CHECK OUT THE w3schools and other tutorials!
Will show example from the quiz show
Task: find questions that player has NOT answered correctly and has NOT been asked that day.

7. Two steps
Create a temporary table of all the questions asked a particular player and answered correctly OR asked today.
need to specify the contents of the table. In this case, one field of INT datatype
If there have been past questions, do a SELECT using LEFT JOIN to extract any question NOT present in the past array. Otherwise, do a simple SELECT

8. $query="CREATE temporary TABLE past (item_id INT)";
1st step: create temporary table, past
$query="CREATE temporary TABLE past (item_id INT)";
$query.= " SELECT question_id FROM history WHERE (player_id='".$player_id;
$query.= "' AND (whenplayed='".$today."' OR correct))";
$result=mysqli_query($link,$query);
$query="SELECT * FROM past";
$result = mysqli_query($link,$query);
$Num_past = mysqli_num_rows($result);

9. 2nd step if ($Num_past>0) {
$sel = "SELECT questions.question_id, question, answerpattern, value from questions";
$sel =$sel . " LEFT JOIN past ON questions.question_id = past.question_id WHERE ";
$sel = $sel . " category='" . $pickedcategory . "' AND past.question_id IS NULL"; }
else {
$sel="SELECT question_id, question, answerpattern, value from questions ";
$sel= $sel . " WHERE category= '" . $pickedcategory. "'";

10. SELECT Conditions WHERE with a JOIN After aggregating using GROUP
Sets condition on individual records
with a JOIN
the ON specifying what field to do the JOIN on, generally a foreign key equal to a primary key
After aggregating using GROUP
HAVING sets a condition on grouped data 10

11. Conditions
Remember: the single equal sign is the operator for equality!
Other comparisons: >, <, >=, <=
LOGIC: AND, OR, NOT
REGEX for regular expressions
LIKE: another way to specify a pattern

12. Conditions Can select using set of values
SELECT * FROM questions WHERE category IN (‘trivia’,’misc’,’silly’)
See also BETWEEN

13. Recall 4 tables In most cases, people have only 1 role.
Affleck is an exception.
In most cases, awards
are for 1 role. Producing
is an exception.
Some roles are not
nominated for anything,
hence the 0.
4 tables
movies
mid
mname
mdate
people
pid
pname …
nominations
aid
rid
category
win …
roles
rid
mid
pid
role (director,actor,etc.)

14. Tasks List all movies by name, ordered by date
SELECT mname, mdate FROM movies ORDER BY mdate
List all people by name with roles in a given movie, named $moviename (this is mixture of php and straight SQL)
SELECT p.pname,m.mname,r.role FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid WHERE m.mname='$moviename'

15. next task: reuse JOIN clauses
List all directors (by name), with movie (by name) ordered by movie name
SELECT p.pname,m.mname FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid WHERE r.role='Director' ORDER BY m.mname

16. List all movies by name in which someone was nominated for Best Lead Actor
SELECT m.mname FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid JOIN nominations as n ON n.rid=r.rid WHERE n.award='Best Lead Actor'

17. next task, again reuse JOIN clauses plus start of WHERE
List all movies by name in which someone was nominated for an acting category. Count number.
Best Lead Actor, Best Lead Actress, Best Supporting Actor, Best Supporting Actress
SELECT m.mname, count(*) FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid JOIN nominations as n ON n.rid=r.rid WHERE n.award IN ('Best Lead Actor, Best Lead Actress, Best Supporting Actor, Best Supporting Actress) ORDER BY m.mname GROUP BY m.mname

18. More
List movie name, person name, nominated award ordered by movie name
SELECT m.mname, p.pname, n.award FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid JOIN nominations as n ON n.rid=r.rid ORDER BY m.mname

19. Next List winners: movie name, person name, award
SELECT m.mname, p.pname, n.award FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid JOIN nominations as n ON n.rid=r.rid ORDER BY m.mname WHERE n.win='true'

20. List movie name, number of people nominated, ordered from high to low
SELECT m.mname, count(*) FROM movies as m JOIN roles as r ON m.mid = r.mid JOIN people as p ON p.pid = r.rid JOIN nominations as n ON n.rid=r.rid ORDER BY m.mname GROUP BY m.mname
EXTRA CREDIT: figure out how not to count multiples for awards that have multiples (such as producer, technical awards)

21. Research Some questions require consideration of absent records
What movies were nominated (had people nominated) for Best Movie but not Best Director ?
Look up and study examples of LEFT JOINS

22. UPDATE Can update a single record or a set of records.
UPDATE questions SET text = ‘$ntext’ WHERE ques_id=‘$qid’
Assumes table names questions with fields text and ques_id and php variables $ntext and $qid set previously

23. UPDATE Raise all the prices 10%
UPDATE products SET price=1.10*price
Raise the prices that are over 100 by 5%
UPDATE products SET price=1.05*price WHERE price > 100

24. Projects
Recall Geolocation / Google maps .
Quizhttp://faculty.purchase.edu/jeanine.meyer/html5/mapmediaquiz.html
Does not use database. Enhancement is to add database for locations

25. Projects Bookmarks, with password system for finders

26. Projects
Student departments

27. Projects
Trivia quiz
edit feature?
change scoring?
improve interface?

28. Projects
Origami store need file upload to upload pictures
general improvement
scale up

29. Projects
Songs and features in songs. Find similar songs

30. Stories

31. Citations

32. Assignment Team project to present and Enhance one of these projects
There will be an assignment to build database project totally on your own

33. Projects Late addition: citations (may need updating) student database
trivia quiz
book marks
origami store
songs
Stories
Google maps media portal (or quiz)
Note: this does NOT have a database at all! So obvious enhancement is to make use of database, possibly with table for player or ???
Late addition: citations (may need updating)

34. Preview (after midterm)
Self assign or I assign teams (3-4)
Teams will pick projects
resolve conflicts

35. Assignment https://donottrack-doc.com/en/
This is a set of short interactive videos concerning tracking. Allow yourself a few days.
Make a response. You may consider including a screen shot on your results.

36. Homework [Prepare to form teams AND choose application.]
Working with existing code is…always a challenge but more common than starting from nothing.
Posting assignment on Do Not Track.