我得到這個代碼的PHP的通知:PHP通知檢索用戶數據
注意:未定義抵消:1
我想到的是功能返回用戶,例如郵件:在這個例子中[email protected]:
$userId = 1;
getUserById($userId, 'mail');
function getUserById($id, $string = '') {
return getUser($string, $id);
}
function getUser($id = '', $string = '') {
if(isLoggedIn() || $id != '') { //If user is logged in or if id is sent
if($string == '') {
return $_SESSION['loggedIn']['fName'] . ' ' . $_SESSION['loggedIn']['lName']; //Return logged in users first name and lastname as a string
}
else {
return $_SESSION['loggedIn'][$string]; //Return logged in users email as a string
}
}
else { //If user is not logged in return false
return false;
}
}
如果我這樣稱呼它:
getUserById($userId, 'phoneNumber');
該函數將返回用戶的電話號碼。
我不知道爲什麼林收到此通知,任何幫助如何解決它是讚賞! :)
您是否開始了會話? –