return 来自数据库的多个值
return multiple values from database
我正在尝试从数据库中回显两个值,并想在 html 中的不同位置回显该值,但是当我在 html 中调用该函数时,它在同一个地方给了我两个值但我想在不同的地方看到这个值,请问如何做到这一点。
HTML
<div id="container">
<div id="header">
<a href="home.php?q=logout">LOGOUT</a>
</div>
<div id="main-body">
<br/><br/><br/><br/>
<h1>
Hello <?php $user->get_fullname($uid); ?>
<?php echo $uid; ?>
</h1>
</div>
<div id="footer"></div>
</div>
php class 名称用户
中的函数
public function get_fullname($uid){
$sql3="SELECT fullname,uemail FROM users WHERE uid = $uid";
$result = mysqli_query($this->db,$sql3);
$user_data = mysqli_fetch_array($result);
echo $user_data['fullname'];
echo $user_data['uemail'];
}
public function get_fullname($uid){
$sql="SELECT fullname,uemail FROM users WHERE uid = $uid";
$result = mysqli_query($this->db,$sql);
$user_data = mysqli_fetch_array($result);
return (object)array('username'=>$user_data['fullname'],'email'=>$user_data['uemail']);
}
/* Call the class method - doesn't echo back to screen */
$userdata=$user->get_fullname($uid);
然后,稍后在代码中:
echo $userdata->username;
/* or */
echo $userdata->email
不过有一件事,也许您应该调查 prepared statements
以便与 mysqli
一起使用
我正在尝试从数据库中回显两个值,并想在 html 中的不同位置回显该值,但是当我在 html 中调用该函数时,它在同一个地方给了我两个值但我想在不同的地方看到这个值,请问如何做到这一点。
HTML
<div id="container">
<div id="header">
<a href="home.php?q=logout">LOGOUT</a>
</div>
<div id="main-body">
<br/><br/><br/><br/>
<h1>
Hello <?php $user->get_fullname($uid); ?>
<?php echo $uid; ?>
</h1>
</div>
<div id="footer"></div>
</div>
php class 名称用户
中的函数 public function get_fullname($uid){
$sql3="SELECT fullname,uemail FROM users WHERE uid = $uid";
$result = mysqli_query($this->db,$sql3);
$user_data = mysqli_fetch_array($result);
echo $user_data['fullname'];
echo $user_data['uemail'];
}
public function get_fullname($uid){
$sql="SELECT fullname,uemail FROM users WHERE uid = $uid";
$result = mysqli_query($this->db,$sql);
$user_data = mysqli_fetch_array($result);
return (object)array('username'=>$user_data['fullname'],'email'=>$user_data['uemail']);
}
/* Call the class method - doesn't echo back to screen */
$userdata=$user->get_fullname($uid);
然后,稍后在代码中:
echo $userdata->username;
/* or */
echo $userdata->email
不过有一件事,也许您应该调查 prepared statements
以便与 mysqli