我得到这个代码的PHP的通知:PHP通知检索用户数据
注意:未定义抵消:1
我想到的是功能返回用户,例如邮件:在这个例子中[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;
}
}
如果我这样称呼它:
getUserById($userId, 'phoneNumber');
该函数将返回用户的电话号码。
我不知道为什么林收到此通知,任何帮助如何解决它是赞赏! :)
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