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php - 从类 php oop 返回值

转载 作者:行者123 更新时间:2023-12-02 06:38:48 25 4
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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);
}
}

现在要获取我这样做的信息:

$info = new userinfo();
$info->fetchdatabyemail('email@email.com');
echo $info->city;

它不返回信息。我认为我做错了任何想法请

最佳答案

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;
}

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.

关于php - 从类 php oop 返回值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12209264/

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