section problems by Sylvia Tashev, Brian Le, and several UW student volunteers (see Acknowledgements section of textbook!)
Today we will solve some practice problems similar to the problems that will be on our final exam. Try these out first using paper and pencil!
HTML / CSS - Apples:
Write HTML (just what's in the body) and CSS code to reproduce the following page:
The following are details about the appearance of the page:
greenapples.jpg
. It also links to http://www.apple.com/.#99cc99
.<div id="main"> <h1>Apples</h1> <p> <q>An apple a day keeps the doctor away!</q><br /> <a href="http://www.apple.com/"><img src="greenapples.jpg" alt="apple" /></a> </p> </div> <div id="about"> <h2>Some types of apples:</h2> <ul> <li>Red delicious</li> <li class="special">Golden delicious</li> <li>Granny Smith</li> <li class="special">Crabapple</li> <li>Gala apple</li> <li class="special">Red Rome apple</li> <li>Ginger gold</li> <li class="special">Fuji apple</li> </ul> <h2>Other apples:</h2> <dl> <dt>Apple iPod</dt> <dd>This apple is poisonous.</dd> <dt>Steve Jobs</dt> <dd>The tallest apple.</dd> </dl> </div>
h1, h2 { color: green; font-family: "Garamond", serif; } img { width: 250px; border: 0px; } q { font-style: italic; } dt { font-weight: bold; } #about, #main { float: left; padding: 20px; } #about { background-color: #99cc99; border: 2px outset green; } .special { font-size: 120%; color: white; }
PHP - Image Gallery Search:
The following HTML snippet is the skeleton of a simple search page for an image gallery.
<h1>Image Gallery Search<h1> <form action="search.php" method="get"> <fieldset> Type a query: <input type="text" name="query" /> <br /> <input type="submit" value="Search" /> </fieldset> </form>
Write a PHP program called search.php
to implement the searching feature. An image is considered a "match" to the search string if the name of the
image contains the entirety of the search string. For example, a query of
"tea" might match "tea.jpg" or "steamboat.jpg".
You should output an unordered list of links to all matches. A blank query should return no results. The gallery stores all of its images in a directory called images
. You may assume there are only image files in the images directory. The search should be case-insensitive and you should eliminate the whitespace surrounding a query before processing it.
Here is a ZIP gallery of images you can use to test your program.
<?php $BASE_URL = "images/"; if (isset($_REQUEST["query"]) && $_REQUEST["query"]) { $query = trim($_REQUEST["query"]); $images = glob($BASE_URL . "*$query*"); if (count($images) > 0) { ?> <ul> <?php foreach ($images as $img) { ?> <li><a href="<?= $img ?>"><?= basename($img) ?></a></li> <?php } ?> </ul> <?php } } ?>
JavaScript - Flower Garden:
Write the necessary JavaScript code garden.js
for the functionality of the following page:
count
".plant
".clear
".daffodils.jpg
or roses.jpg
.Here is a runnable solution.
document.observe("dom:loaded", function() { $("plant").observe("click", plant); $("clear").observe("click", clear); }); // adds flower patches to the garden (randomly picking daffodils or roses) // the count of how many is taken from the text box function plant() { var count = $("count").value; for (var i = 0; i < count; i++) { var patch = document.createElement("img"); // pick random flower patch var n = Math.random().round(); var type = "daffodils"; if (n == 1) { type = "roses"; } patch.src = type + ".jpg"; patch.alt = type; $("garden").appendChild(patch); } } // removes all the flower patches so we can start planting again! function clear() { var patches = $$("#garden img"); for (var i = 0; i < patches.length; i++) { patches[i].remove(); } }
Ajax/JavaScript - Music Search:
Write the necessary JavaScript code for the Ajax functionality of the following page.
This page allows the user to search for songs. You are given access to a web service music.php
. To use the service, you may send in a query parameter named term
. The service will then return all of the songs that contain the search term in the following XML format.
If you send the empty string '', the service will return a list of all songs.
<songs> <song genre="Pop" duration="3:48"> <title>Dance Like Michael Jackson</title> <artist>The Far East Movement</artist> <album>Animal</album> </song> <song genre="Soundtrack" duration="3:11"> <title>Aaj Ki Raat</title> <artist>A. R. Rahman</artist> <album>Slumdog Millionaire</album> </song> </songs>You will need to parse the XML to display all the songs returned to the page.
search
"query
"results
".onFailure
or onException
functions.document.observe("dom:loaded", function() { $("search").observe("click", search); }); // makes an ajax request to the server for the list of songs that satisfy the given query function search() { new Ajax.Request("music.php?term=" + $("query").value, { method: "get", onSuccess: showList, }); } // puts the results into the page function showList(ajax) { // clear previous results, if any $("results").innerHTML = ""; // display song data var songs = ajax.responseXML.getElementsByTagName("song"); for (var i = 0; i < songs.length; i++) { var title = songs[i].getElementsByTagName("title")[0].firstChild.nodeValue; var artist = songs[i].getElementsByTagName("artist")[0].firstChild.nodeValue; var duration = songs[i].getAttribute("duration"); var song = document.createElement("p"); song.innerHTML = title + " - " + artist + " - " + duration; $("results").appendChild(song); } if (songs.length == 0) { $("results").innerHTML = "no results"; } }
SQL - Duplicate Cities:
Some countries have two cities with the same name. For example, there are several cities in the USA named Springfield, which is part of the reason that name was chosen for the home town of the Simpsons.
Using the world
database, find all such cities where there are two or more cities within the same country that have the same name. Show both the country name and the city name. Eliminate duplicate results and order by the country name, breaking ties by the city name.
Recall the world
database schema:
code | name | continent | independence_year | population | gnp | head_of_state | ... |
---|---|---|---|---|---|---|---|
AFG | Afghanistan | Asia | 1919 | 22720000 | 5976.0 | Mohammad Omar | ... |
NLD | Netherlands | Europe | 1581 | 15864000 | 371362.0 | Beatrix | ... |
... | ... | ... | ... | ... | ... | ... | ... |
id | name | country_code | district | population |
---|---|---|---|---|
3793 | New York | USA | New York | 8008278 |
1 | Los Angeles | USA | California | 3694820 |
... | ... | ... | ... | ... |
country_code | language | official | percentage |
---|---|---|---|
AFG | Pashto | T | 52.4 | NLD | Dutch | T | 95.6 |
... | ... | ... | ... |
This is the output you should get:
+--------------------+--------------+ | country | city | +--------------------+--------------+ | China | Jining | | China | Jinzhou | | China | Kaiyuan | | China | Suzhou | | China | Yichun | | Indonesia | Depok | | Mexico | Guadalupe | | Mexico | La Paz | | Mexico | Matamoros | | Philippines | San Carlos | | Philippines | San Fernando | | Philippines | San Jose | | Russian Federation | Zeleznogorsk | | United States | Arlington | | United States | Aurora | | United States | Columbus | | United States | Glendale | | United States | Kansas City | | United States | Pasadena | | United States | Peoria | | United States | Richmond | | United States | Springfield | +--------------------+--------------+ 22 rows in set (14.48 sec)
SELECT DISTINCT u.name as country, c1.name as city FROM cities c1 JOIN countries u ON u.code = c1.country_code JOIN cities c2 ON c2.country_code = c1.country_code WHERE c1.name = c2.name AND c1.id <> c2.id ORDER BY u.name, c1.name;