我想知道我哪里错了.

[chilychang]

<?php
function get_director($director_id)
{
global $db;
$query='SELECT people_fullname FROM people WHRER people_id='.$director_id;
$result=mysql_query($query,$db);
$row=mysql_fetch_assoc($result);
extract($row);
return $people_fullname;
}

function get_leadactor($leadactor)
{
global $db;
$query='SELECT people_fullname FROM people WHRER people_id='.$leadactor;
$result=mysql_query($query,$db);
$row=mysql_fetch_assoc($result);
extract($row);
return $people_fullname;

}

function get_movietype($type_id)
{
global $db;
$query='SELECT movietype_label FROM movietype WHERE movietype_id='.$type_id;
$result=mysql_query($query,$db);
$row=mysql_fetch_assoc($result);
extract($row);
return $movietype_label;
}

$db=mysql_connect("localhost","root","xiaozhuo") or die ('Unable to connect MYSQL');
mysql_select_db('moviesite',$db) or die (mysql_error($db));

$query='SELECT movie_name,movie_year,movie_director,movie_leadactor,movie_type FROM movie ORDER BY movie_name ASC,movie_year DESC';
$result=mysql_query($query,$db) or die (mysql_error($db));

$num_movies=mysql_num_rows($result);

$table=<<<ENDHTML
<div style="text-align:center;">
<h2>Movie Review Database</h2>
<table border="1" cellpadding="2" cellspacing="2" style="width:70%; margin-left:auto; margin-right:auto;">
<tr>
<th>Movie Title</th>
<th>Year of Release</th>
<th>Movie Director</th>
<th>Movie Lead Actor</th>
<th>Movie Type</th>
</tr>
ENDHTML;
while ($row=mysql_fetch_assoc($result))
{
extract($row);
$director=get_director($movie_director);
$leadactor=get_leadactor($movie_leadactor);
$movietype=get_movietype($movie_type);

$table.=<<<ENDHTML
<tr>
<td>$movie_name</td>
<td>$movie_year</td>
<td>$movie_director</td>
<td>$movie_leadactor</td>
<td>$movie_type</td>
</tr>
ENDHTML;
}
$table.=<<<ENDHTML
</table>
<p>$num_movies Movies</p>
</div>
ENDHTML;
echo $table;
?>

[chilychang]

难道分低了.?没人帮忙.?

[chilychang]

运行以后显示这样的错误

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\table2.php on line 7

Warning: extract() [function.extract]: First argument should be an array in C:\AppServ\www\table2.php on line 8

...

[do_fork]

判断mysql的返回值先

[chilychang]

这个要怎么改呀.?

[do_fork]

$result=mysql_query($query,$db);
$row=mysql_fetch_assoc($result);

所有的result，必须先判断后使用

[qtffly]

看来看去，数据库部分代码的语法总是很别扭.

[chilychang]

-0-我才学,,才学,,要谅解,,

[mrshelly]

$result=mysql_query($query,$db) or die('Error:'.mysql_error()." SQL:{$query}");

学会调试信息.以及代码的严谨