如何將數據庫中的SELECTED元素放入變量

Question

如何將這些元素放置到變量中？

$user_information = $connect->prepare("SELECT email, gender, firstname, middlename, FROM members"); 

像

$firstname = 'firstname'; 
$gender = 'gender'; 
$email = 'email'; 

我怎麼能提前

Answer 1

做到這一點

謝謝如果你正在使用PHP和PDO，你可以在獲取方法使用，以獲得一個使用PDO::FETCH_OBJ的對象。 http://de3.php.net/manual/en/pdostatement.fetch.php

$result = $sth->fetch(PDO::FETCH_OBJ); 
print $result->NAME; 

Answer 2

list($email, $gender, $firstname, $middlename) = $user_information; 

欲瞭解更多信息檢查LIST function @ PHP docs。

Answer 3

如果你是，如果你使用的mysqli

$user_information->execute(); 
$user_information->bind_result($firstname, $gender, $email); 
$user_information->fetch(); 
printf("%s, %s, %s\n", $firstname, $gender, $email); 
$user_information ->close(); 

Answer 4

如果它返回一個對象，你可以得到它喜歡用PDO

$user_information ->execute(); 
    $result = $user_information->fetch(PDO::FETCH_ASSOC); 
    $firstname = $result['firstname']; 
    $gender = $result['gender']; 
    $email = $result['email']; 

;

$firstname = $user_information->firstname; 
$gender = $user_information->gender; 
$email = $user_information->email; 

或者如果它返回一個數組;

$firstname = $user_information["firstname"]; 
$gender = $user_information["gender"]; 
$email = $user_information["email"];