所以我从PHP数据库中获取数据。我打算比较从查询中获得的数据创建的每个对象的值。到现在为止还挺好。如何从PHP中的值对象获取数据?
当我尝试从对象中检索数据后,就会出现这个问题,一旦它们被创建,它就变成空的。
我不是PHP开发人员,所以我不知道我是否遵循适当的PHP逻辑。我习惯了JS,AS3和Java,所以PHP中的对象和值对象有点不同于我所知道的。
任何人都知道我如何检索我的数据?
<?php
include("../config.php");
class userVO
{
public $uid;
public $name;
public $email;
public $list;
public $num_list_items;
public $matches;
public $num_matches;
public function __construct()
{
$this->matches = array();
}
}
mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die(mysql_error());
mysql_select_db(DB_NAME) or die(mysql_error());
$json_array = array();
$result = mysql_query("...");
$num_results = mysql_numrows($result);
mysql_close();
$users = array();
$i = 0;
while($i < $num_results)
{
$match = new userVO;
$match->uid = mysql_result($result, $i, "uid");
$match->name = mysql_result($result, $i, "name");
$match->email = mysql_result($result, $i, "email");
$users[] = userVO;
$i++;
}
$num_users = count($users);
echo "num users: " . $num_users . "<br>";
$i = 0;
while($i < $num_users)
{
echo "--- i: " . $i . " ---<br>";
$current_user = $users[$i];
echo "users[" . $i . "]: " . $users[$i] . "<br>";
echo "users[" . $i . "]->name: " . $users[$i]->name . "<br>";
echo "current user: " . $current_user . "<br>";
echo "current user name: " . $current_user->name . "<br>";
$i++;
}
?>
也许你的意思'$用户[] = $匹配;'? – 2011-02-25 22:36:37