Value outside of function is different from value inside function - php

Question

In this elementary code, I check if a record exists with a particular itemId for a particular user and country.
The record returns 1 inside the function, but when it's outside, it's something totally different.
function recordExists(&$d) {
global $conn, $userId, $userCountryCode;
$sqlFindRecord = "select count(itemId) as recordCount from onr_items_mod where itemTempId = :itemTempId and itemStatus IN(0,5,8) and itemPublishedStatus = 2 and itemUserId = :userId and itemUserCountryCode = :userCountryCode";
$countRecord = $conn->prepare($sqlFindRecord);
$countRecord->execute(array(
":itemTempId" => $d,
":userId" => $userId,
":userCountryCode" => $userCountryCode
));
$countRecord->bindColumn('recordCount',$d);
echo 'This is D '.$d.' ---';
$row = $countRecord->fetch();
if($d == 0) {
return $d;
} elseif ($d == 1) {
return $d;
} else {
return false;
}
}
$d = 'US01'
if(recordExists($d) == 0) { //If the function returned 0
echo $d;
echo 'Was Zero';
} elseif (recordExists($d) == 1) { //If the function returned 1
echo $d;
echo 'Was One';
} else {
echo $d;
echo 'Neither Zero or One';
}
After I do the above, it's always 'Neither Zero or One'. Why is this? I tried the entire morning, but just cannot figure a way to sort this out.
This works, but not the above
function counter(&$a) {
if($a == 0) {
return $a;
} elseif ($a == 1) {
return $a;
} else {
return false;
}
}
$a = 1;
if(counter($a) == 0) {
echo $a;
} elseIf (counter($a) == 1) {
echo $a;
} else {
echo 'Umm...'.$a;
}

Answer 1

It will always be 'Neither Zero or One' why? Let's seee how it works
1. $d = 'US01'
2. if(recordExists($d) == 0) { //If the function returned 0
3. echo $d;
4. echo 'Was Zero';
5. } elseif (recordExists($d) == 1) { //If the function returned 1
6. echo $d;
7. echo 'Was One';
8. } else {
9. echo $d;
10. echo 'Neither Zero or One';
11. }
First $d is equal US01. It's ok. But after pass line #2 $d == 1 because you have set it in function (you have pass $d by reference).
So if statement in #2 line will return false beacuse thera are some records in database with itemTempId = 'US01'.
But now in the second if statement in line #5 variable $d = 1. So I suposse that you don't have any record with that itemTempId in your database. So the second statement (line #5) will also return false.
In my opinion you shouldn't pass it by reference.
EDIT
If you want to check if item exist in databse then yopu can ask only once. I really don't know what your goal is. But maybe this will be good
$d = 'US01';
$ret = recordExists($d);
if ($ret === 0) {
echo 'Was zero';
} elseif ($ret === 1) {
echo 'Was one';
} else {
echo 'Something else';
}

Answer 2

try to insert that into session variable and retrieve it in another method.
Hope it'll help.