object existance not being recognised in PHP - php

I have the following script:
function canLogin($mail, $password) {
if( isset($mail, $password) ) {
$login = dbFind('users', array("mail" => (string)"$mail", "pass" => (string)"$password"), true);
var_dump($login);
if($login)
{ return (string)$login['_id']; }
else { return false; }
} else { return false; }
}
which function dbFind is the following:
function dbFind($collIn, $obj, $one = false) {
global $db;
$collection = $db->$collIn;
if($one == true)
return $collection->findOne($obj);
else return $collection->find($obj);
}
now, as you can notice from the first snippet of code, I am using var_dump to see the content of the $login variable and afterwards check if it is set or not.
When I have a right login, the var_dump returns the correct object from the db, otherwise it returns NULL. The problem is that in the if statement just after the var_dump always returns false!
What I am missing? Thanks in advance.

If (isset($login)){
//do what you need
}

if($login != NULL){
return (string)$login['_id'];
}else{
return false;
}
I had such an issue and this worked and it's working.

Try changing the if statement to something like:
if (is_array($login) && array_key_exists('_id', $login) && $login['_id'] instanceof MongoId)
{
//the array_key_exists check could be omitted, but is there just in case
return (string) $login['_id'];
}
You can omit the first two checks and use !empty($login['_id']) && $login['_id'] instanceof MongoId, but that might issue warnings (not sure).BTW: try setting your ini file to E_STRICT | E_ALL, maybe that'll give you some more clues as to what's going wrong.
From what you've posted in your comment, and looking at your code, this line: return (string)$login['_id']; is actually casting an object to a string. The result looks like this:
array(5) { ["_id"]=> object(MongoId)#8 (1) //<-- OBJECT
{ ["$id"]=> string(24) "50d19fed9cc2318521000001" }
}
And you're using the (string) cast, that just doesn't feel right As Sammaye pointed out, that should work. You're casting too much BTW: "$foo" casts $foo to a string, because of the double quotes, no need for an extra cast... but that's not the point... Anyway, try this:
return $login['id']->{'$id'};//need the single quotes, the property NAME is $id, not id
See the docs on the mongoid class for more details on the methods/properties, and headaches...
PS: it might be cleaner to use the getter method:
$login['_id']->getPID();//returns an int, not a string, though
Good luck, happy coding

Related

PHP array value isn't shown by var_dump but it was fetched.

I wrote some program to check the value in an array.
var_dump($profileuser);//NULL
$profileuser = get_user_to_edit($user_id);//fetch the value of $profileuser
var_dump($profileuser);//does not print the value of $profileuser->user_url
//nor by print_r($profileuser)
if(isset($profileuser->user_url))
echo $profileuser->user_url;//printed!!!!How is it possible??
Could somebody can explain how this happened?
background:
I modified the kernel of wordpress.
This happened when I modified the file of wp-admin/user-edit.php.
You say it's an array, but you're accessing it as an object ($obj->foo rather than $arr['foo']), so it's most likely an object (actually it is - get_user_to_edit returns a WP_User). It could easily contain the magic __get and __isset methods that would lead to this behaviour:
<?php
class User {
public $id = 'foo';
public function __get($var) {
if ($var === 'user_url') {
return 'I am right here!';
}
}
public function __isset($var) {
if ($var === 'user_url') {
return true;
}
return false;
}
}
$user = new User();
print_r($user);
/*
User Object
(
[id] => foo
)
*/
var_dump( isset($user->user_url) ); // bool(true)
var_dump( $user->user_url ); // string(16) "I am right here!"
DEMO
One possibility is that $profileuser is an Object that behaves as an array and not an array itself.
This is possible with interface ArrayAccess. In this case, isset() would return true and you might not see it when you do var_dump($profileuser);.
When you want an object to behave like an array, you need to implement some methods which tell your object what to do when people use it as if it were an array. In that case, you could even create an Object that, when accessed as an array, fetches some webservice and return the value. That may be why you are not seeing those values when you var_dump your variable.
I thinks it's not possible, I've created test code and var_dump behaves correctly. Are you 100% sure you don't have any typo in your code? I remind variables in PHP are case sensitive
<?php
$profileuser = null;
class User
{
public $user_url;
}
function get_user_to_edit($id) {
$x = new User();
$x->user_url = 'vcv';
return $x;
}
var_dump($profileuser);//NULL
$user_id = 10;
$profileuser = get_user_to_edit($user_id);//fetch the value of $profileuser
var_dump($profileuser);//does not print the value of $profileuser->user_url
//nor by print_r($profileuser)
if(isset($profileuser->user_url))
echo $profileuser->user_url;//printed!!!!How does it possible??
Result is:
null
object(User)[1]
public 'user_url' => string 'vcv' (length=3)
vcv

PHP create object inside it's own constructor

EDIT: Link to my first question. Might clear some things up.
PHP Get corresponding data, with default and error handling
I have a function that checks if a GET statement exists. If so, it passes the value to a other function that then selects a class based on the value of the GET statement.
explaining:
The url = Page=Contact
The GetFormVariable approves it, and the class Contact is selected and it will give back a string. This string is used as an object 'Content' that, as it says, creats the content of the page.
public function getFormVariable($value){
switch (strtoupper($_SERVER['REQUEST_METHOD'])) {
case 'GET':
if (isset($_GET[$value]) && $_GET[$value] != NULL) {
return $_GET[$value];
}
else{
return false;
}
break;
case 'POST':
if (isset($POST[$value]) && $POST[$value] != NULL) {
return $POST[$value];
}
else{
return false;
}
break;
default:
return false;
}
}
Now the question.
When there is no GET statement in the url. The GetFormVariable returns false. And this means there is nothing shown.
How do i give this constructor.
public function SetProperty ($prob, $val){
$this->$prob = $val;
}
The information to create the ContentHome.
SetProperty('Content', 'ContentHome');
Sorry for poor explanation, if anything is unclear please tell me so.
I'm suggesting we close this question as unclear what you're asking, but decided to throw some help on the provided code sample anyway...
You can strip this down dramatically, and since there's no context, the function can be static too.
static public function getFormVariable($value)
{
if($_SERVER['REQUEST_METHOD'] == 'GET' &&
isset($_GET[$value]) &&
!empty($_GET[$value]))
return $_GET[$value];
elseif($_SERVER['REQUEST_METHOD'] == 'POST' &&
isset($POST[$value]) &&
!empty($POST[$value]))
return $POST[$value];
return false;
}
Your original isset and != NULL checks were doing the same check. Maybe you want the empty() check as a third check, but look it up to be certain.
The question is unclear.
How you call getFormVariable? How you use SetProperty with the information getFormVariable provides?
As far as I understood, you mean this...
$var = getFormVariable(???);
if (false === $var)
{
SetProperty('Content', 'ContentHome');
} else {
SetProperty('var', $var);
}

PHP return the value but false

I have this simple function:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status']))
return $status['status'];
return false;
}
Now I am looking for a way to return false yes, but to return also the value of the variable.
I tried the following, but it makes it empty anyway:
return $status['status'] == false;
So the logi is return false anyway but give me back also the value of the variable, even if it's false, because false should not mean empty :)
Return an array, and use the list() method to get your results:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
$statusString = $status['status'];
$statusFlag = false;
if(isAllowed($status['status']))
$statusFlag = true;
return array($statusFlag,statusString);
}
//usage
list($flag,$msg) = isMember(5,"whatever");
echo "Access: $flag, with message $msg";
A function can not return multiple values, but similar results can be obtained by (1) returning an array or by (2) passing a variable by reference and storing the value you want returned in that variable.
You will need to write your function in a way that it returns an array containing the following:
The value you wan't returned
A flag that signifies true/false
Pass a variable by reference into your function and store the value of the status in that variable.
function isMember($uID, $pdo, &statByRef) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status'])) {
return $status['status'];
}
$statByRef = $status['status'];
return false;
}
False returns empty in PHP, see http://php.net/manual/en/function.empty.php
From documentation:
Determine whether a variable is considered to be empty. A variable is considered empty if it does not exist or if its value equals FALSE. empty() does not generate a warning if the variable does not exist.
Try using something like this:
function isMember($uID, $pdo) {
$status = getUserStatus($uID, $pdo);
if(isAllowed($status['status'])){
return $status['status'];
}
return false;
} // If isAllowed returns true, then PHP will return $Status['Status'];, if not then PHP will by default return false.
I have noticed you haven't used braces which makes the code a little awkward to debug. Then validate like:
if (isMember($Var,$AnotherVar) !== false){
//isMember has not returned false, so PHP is operating within these braces
}
Such a simple thing, which should be most effective.
If your wanting to assign true/false to $status['status']; then you are performing the right method, but wrong operator.
== is a comparision operator. Not assignment
= is an assignment operator, so your assignment should be:
$status['status'] = false;

returning from multiple points in a function

This is more or less a readability, maintainability and/or best practice type question.
I wanted to get the SO opinion on something. Is it bad practice to return from multiple points in a function? For example.
<?php
// $a is some object
$somereturnvariable = somefunction($a);
if ($somereturnvariable !== FALSE) {
// do something here like write to a file or something
}
function somefunction($a) {
if (isset($a->value)) {
if ($a->value > 2) {
return $a->value;
} else {
return FALSE;
} else {
// returning false because $a->value isn't set
return FALSE;
}
}
?>
or should it be something like:
<?php
// $a is some object
$somereturnvariable = somefunction($a);
if ($somereturnvariable !== false) {
// do something here like write to a file or something
}
function somefunction($a) {
if (isset($a->value)) {
if ($a->value > 2) {
return $a->value;
}
}
return FALSE
}
?>
As a matter of practice, I always try to return from ONE point in any function, which is usually the final point. I store it in a variable say $retVal and return it in the end of the function.It makes the code look more sane to me.
Having said that, there are circumstances where say, in your function as the first line you check if a var is null and if yes you are returning. In this case, there is no point in holdin on to that variable, then adding additional checks to skip all the function code to return that in the end.
So...in conclusion, both ways works. It always depends on what the situation is and what you are more comfortable with.

Function not returning

function checkSessionToken($token) {
if (!isset($_SESSION['token']) || empty($_SESSION['token'])) {
return false;
}
return $token == $_SESSION['token'];
}
For some reason this function is not returning any value, while it really should, causing my script to fail. When I put the result in a variable and echo it, I don't get any output. Why is this?
Thank you
The code looks good.
Don't echo the value. Use var_dump. echo false; does not output anything (example).
var_dump will output bool(false).
return $token == $_SESSION['token']; may return NULL if there is no return value because of one of your data. You can check it with print gettype(checkSessionToken($token));.
To avoid this you could make sure to return the bool result of (foo == bar), e.g. adding parenthesis as per follow: return ($token == $_SESSION['token']);

Categories