2019-02-21 11:07:30 +01:00
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<title>Einführung in Linux</title>
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<meta name="description" content="YALC - Yet Another Linux Course ">
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<meta name="author" content="Daniel Schubert">
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<section data-transition="slide" data-background="#4d7e65" data-background-transition="zoom">
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<h1>LINUX</h1>
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<h3>Das universelle Betriebssystem</h3>
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<p>Eine Einführung - Teil 14
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<p> <small>Deutsche Angestellten Akademie</small> </p>
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<aside class="notes">
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Frage: Fragen zur letzten Stunde?
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</aside>
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</section>
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<section data-transition="slide" data-background="#b5533c" data-background-transition="convex"> <h3>lampp stack</h3> </section>
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2019-02-28 09:57:43 +01:00
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2019-02-21 11:07:30 +01:00
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<section>
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<pre><code class="bash">~$ sudo apt install apache2 mysql-server phpmyadmin</code></pre>
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<pre><code class="bash">~$ mysql -u root -p
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MariaDB [(none)]> CREATE USER 'newuser'@'localhost' IDENTIFIED BY 'some_very_complex_password';
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MariaDB [(none)]> GRANT ALL PRIVILEGES ON * . * TO 'newuser'@'localhost'; FLUSH PRIVILEGES;
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MariaDB [(none)]> \q
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</code></pre>
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<ul>
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<li class="fragment ">http://localhost/phpmyadmin</li>
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<li class="fragment ">user: phpmyadmin</li>
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<li class="fragment ">http://localhost/phpmyadmin</li>
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</ul>
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</section>
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2019-02-28 09:57:43 +01:00
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<section>
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<h3>mysql</h3>
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</section>
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<section>
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mysql ist eine „relationale Datenbank”
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<p><cite> Relationale Datenbanken versuchen die Realität in einem Datenmodell abzubilden.</cite>
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<ul>
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<li>Jede Zeile einer Tabelle ist ein Datensatz</li>
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<li>Jede Zeile ( Tupel ) besteht aus Attributen ( die Spalten )</li>
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<li>Über Primärschlüssel werden Beziehungen hergestellt</li>
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<li>Eine Datenbank kann viele Tabellen beinhalten</li>
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</ul>
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<img src="img/relationales_datenbankmodell_beispiel.jpg">
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<p><small style="font-size:.4em">Quelle: http://www.datenbanken-verstehen.de/datenbank-grundlagen/datenbankmodell/relationales-datenbankmodell/</small>
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</section>
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<section>
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<h4>Normalisierung</h4>
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Aufteilung der Attribute (Spalten) in mehrere Relationen ( Tabellen ), so dass eine Form ohne Redundanzen entsteht.
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</section>
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<section>
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<img src="img/Aufspaltung_einer_Tabelle_in_zwei_(Beispiel).svg">
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</section>
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<section>
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<img src="img/Wp_Normal_1_301208_165800.JPG">
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</section>
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<section>
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Verbindung aufnehmen:
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<pre><code class="bash">~$ sudo mysql
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# oder:
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~$ mysql -u BENUTZERNAME -p -h localhost
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</code></pre>
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</section>
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<section>Datenbank erstellen
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<pre><code class="sql">CREATE DATABASE IF NOT EXISTS test;
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USE test;
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CREATE TABLE IF NOT EXISTS books (
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BookID INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
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Title VARCHAR(100) NOT NULL,
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SeriesID INT, AuthorID INT);
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CREATE TABLE IF NOT EXISTS authors
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(id INT NOT NULL PRIMARY KEY AUTO_INCREMENT);
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CREATE TABLE IF NOT EXISTS series
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(id INT NOT NULL PRIMARY KEY AUTO_INCREMENT);
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INSERT INTO books (Title,SeriesID,AuthorID)
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VALUES('The Fellowship of the Ring',1,1),
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('The Two Towers',1,1), ('The Return of the King',1,1),
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('The Sum of All Men',2,2), ('Brotherhood of the Wolf',2,2),
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('Wizardborn',2,2), ('The Hobbbit',0,1);</code></pre>
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</section>
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<section>Tabellen anzeigen
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<pre><code class="sql">
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SHOW TABLES;
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+----------------+
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| Tables_in_test |
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+----------------+
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| authors |
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| books |
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| series |
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+----------------+
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3 rows in set (0.00 sec)
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</code></pre>
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</section>
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<section>
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<pre><code class="sql">DESCRIBE books;
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+----------+--------------+------+-----+---------+----------------+
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| Field | Type | Null | Key | Default | Extra |
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+----------+--------------+------+-----+---------+----------------+
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| BookID | int(11) | NO | PRI | NULL | auto_increment |
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| Title | varchar(100) | NO | | NULL | |
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| SeriesID | int(11) | YES | | NULL | |
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| AuthorID | int(11) | YES | | NULL | |
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+----------+--------------+------+-----+---------+----------------+
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</code></pre>
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</section>
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<section>
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Daten anzeigen:
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<pre><code class="sql">SELECT * FROM books;
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+--------+----------------------------+----------+----------+
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| BookID | Title | SeriesID | AuthorID |
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+--------+----------------------------+----------+----------+
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| 1 | The Fellowship of the Ring | 1 | 1 |
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| 2 | The Two Towers | 1 | 1 |
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| 3 | The Return of the King | 1 | 1 |
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| 4 | The Sum of All Men | 2 | 2 |
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| 5 | Brotherhood of the Wolf | 2 | 2 |
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| 6 | Wizardborn | 2 | 2 |
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| 7 | The Hobbbit | 0 | 1 |
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+--------+----------------------------+----------+----------+
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7 rows in set (0.00 sec)</code></pre></section>
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<section>Daten einfügen:
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<pre><code class="sql">INSERT INTO books (Title, SeriesID, AuthorID)
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VALUES ("Lair of Bones", 2, 2);
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Query OK, 1 row affected (0.00 sec)</code></pre>
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<pre><code>SELECT * FROM books;</code></pre></section>
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<section>
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Daten modifizieren:
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<pre><code class="sql">UPDATE books
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SET Title = "The Hobbit"
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WHERE BookID = 7;
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Query OK, 1 row affected (0.00 sec)
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Rows matched: 1 Changed: 1 Warnings: 0</code></pre>
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</section>
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2019-02-21 11:07:30 +01:00
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<!--
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<section>
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Übungen
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C 11.11 [3] Besorgen Sie sich eine geeignete Datei (etwa ein größeres tar -Ar-
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chiv mit Quellcode, vielleicht für den Linux-Kernel) und komprimieren Sie
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sie jeweils mit gzip , bzip2 und xz . Messen Sie dabei die Ausführungszeit
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durch ein vorgesetztes time . Wie verhalten die Laufzeiten und die Kompri-
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mierungsraten sich zueinander?
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C 11.12 [2] Experimentieren Sie mit der Datei aus der vorigen Aufgabe und
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vergleichen Sie die Laufzeit und die Größe des Resultats für xz mit verschie-
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denen Effizienz-Optionen. Was ist der Unterschied zwischen -1 und -6 und
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zwischen -6 und -9 ?
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</section>
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-->
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