代码如下:
Class userinfo {
function fetchdatabyemail($email) {
$result=mysql_query(" SELECT * FROM users WHERE email='$email'");
while($row = mysql_fetch_array($result)) {
$name = $row['name'];
$num = $row['num'];
$city = $row['city'];
}
$numrows= mysql_num_rows($result);
}
}
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现在获取信息,我这样做:
$info = new userinfo();
$info->fetchdatabyemail('email@email.com');
echo $info->city;
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并且它不返回信息。我想我做错了任何主意
做吧
public $numrows;
public function fetchDataByEmail($email) {
$result=mysql_query(" SELECT * FROM users WHERE email='$email'");
while($row = mysql_fetch_assoc($result)) {
$fetch[] = $row;
}
$this->numrows = mysql_num_rows($result);
return $fetch;
}
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然后
$info = new userinfo();
$detail = $info->fetchDataByEmail('email@email.com');
print_r($detail); // return all result array
$info->numrows; // will return number of rows.
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