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Проблем с MySQL


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#1 4e4a

4e4a

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Публикувано: 08 January 2010, 00:12

Здравейте, работя с една книга за програмиране на PHP5, Apache и MySQL и стигнах до едно място, където трябва да се създадат 2 файла createmovie.php и moviedata.php. Ето кодовете на createmovie.php и съответно на moviedata.php:

<?php
//connect to MySQL; note we’ve used our own parameters- you should use
//your own for hostname, user, and password
$connect = mysql_connect("localhost", "Chavdar", "12345") or
die ("Hey loser, check your server connection.");

//create the main database if it doesn't already exist
$create = mysql_query("CREATE DATABASE IF NOT EXISTS moviesite")
or die(mysql_error());

//make sure our recently created database is the active one
mysql_select_db("moviesite");

//create "movie" table
$movie = "CREATE TABLE movie (
movie_id int(11) NOT NULL auto_increment,
movie_name varchar(255) NOT NULL,
movie_type tinyint(2) NOT NULL default 0,
movie_year int(4) NOT NULL default 0,
movie_leadactor int(11) NOT NULL default 0,
movie_director int(11) NOT NULL default 0,
PRIMARY KEY (movie_id),
KEY movie_type (movie_type,movie_year)
)";

$results = mysql_query($movie)
or die (mysql_error());

//create "movietype" table
$movietype = "CREATE TABLE movietype (
movietype_id int(11) NOT NULL auto_increment,
movietype_label varchar(100) NOT NULL,
PRIMARY KEY (movietype_id)
)";

$results = mysql_query($movietype)
or die(mysql_error());

//create "people" table
$people = "CREATE TABLE people (
people_id int(11) NOT NULL auto_increment,
people_fullname varchar(255) NOT NULL,
people_isactor tinyint(1) NOT NULL default 0,
people_isdirector tinyint(1) NOT NULL default 0,
PRIMARY KEY (people_id)
)";

$results = mysql_query($people)
or die(mysql_error());

echo "Movie Database successfully created!";

?>

<?php
//connect to MySQL
$connect = mysql_connect("localhost", "Chavdar", "12345")
or die ("Hey loser, check your server connection.");

//make sure we're using the right database
mysql_select_db("moviesite");

//insert data into "movie" table
$insert = "INSERT INTO movie (movie_id, movie_name, movie_type, " .
"movie_year, movie_leadactor, movie_director) " .
"VALUES (1, 'Bruce Almighty', 5, 2003, 1, 2), " .
"(2, 'Office Space', 5, 1999, 5, 6), " .
"(3, 'Grand Canyon', 2, 1991, 4, 3)";
$results = mysql_query($insert)
or die(mysql_error());

//insert data into "movietype" table
$type = "INSERT INTO movietype (movietype_id, movietype_label) " .
"VALUES (1,'Sci Fi'), " .
"(2, 'Drama'), " .
"(3, 'Adventure'), " .
"(4, 'War'), " .
"(5, 'Comedy'), " .
"(6, 'Horror'), " .
"(7, 'Action'), " .
"(8, 'Kids')" ;
$results = mysql_query($type)
or die(mysql_error());

//insert data into "people" table
$people = "INSERT INTO people (people_id, people_fullname, " .
"people_isactor, people_isdirector) " .
"VALUES (1, 'Jim Carrey', 1, 0), " .
"(2, 'Tom Shadyac', 0, 1), " .
"(3, 'Lawrence Kasdan', 0, 1), " .
"(4, 'Kevin Kline', 1, 0), " .
"(5, 'Ron Livingston', 1, 0), " .
"(6, 'Mike Judge', 0, 1)";
$results = mysql_query($people)
or die(mysql_error());

echo "Data inserted successfully!";
?>

Когато ги стартирам, както пише в книгата, не се зарежда нищо, а дава само бял екран, при положение, че съм създал user с име Chavdar и парола 12345 в MySQL:

mysql> create user 'Chavdar'@'localhost' identified by '12345';
Query OK, 0 rows affected (0.13 sec)

mysql> set password for 'Chavdar'@'localhost' = password('12345');
Query OK, 0 rows affected (0.00 sec)

Някой има ли някакви идеи защо се получава така и как да се оправи този проблем?
Благодаря предварително!




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