PHP FetchAll() Not returning rows - php

Question

I was wondering if somebody could help shed some light as to why this PHP code is not entering into the for loop? In MySQL the query returns the appropriate rows that I need but in this PHP file it fails to return anything into the array, thus not executing the foreach loop.
CODE
<?php
try {
$sql = 'SELECT FirstName,LastName ';
$sql .= 'FROM adventureworks.employeedepartmenthistory ';
$sql .= 'JOIN employee USING(EmployeeID) ';
$sql .= 'JOIN contact USING(ContactID) ';
$sql .= 'WHERE DepartmentID = 2';
$stmt = $dbh->prepare($sql);
$stmt->execute();
echo $sql;
echo '<ul>';
foreach ($stmt->fetchAll() as $emps){
echo "<li>" . $emps["FirstName"] . $emps["LastName"] . "</li>";
}
echo '</ul>';
$stmt = null;
}
catch (Exception $e) {
echo "Error";
echo $e->getMessage();
}
?>
I even have the program echo the SQL query and I copy that into MySQL and it still works. Could it be a simple syntax error or is it the table joins that im performing?
Also im certain that the program is contacting the database correctly because I have other similar PHP files working properly like this one:
<?php
try {
$sql = 'SELECT department.Name FROM adventureworks.department';
$stmt = $dbh->prepare($sql);
$stmt->execute();
echo '<ul>';
foreach ($stmt->fetchAll() as $depts) {
echo "<li>" . $depts["Name"] . " -> (" .
"<a href='deptEmps.php?deptID=" . $depts['deptID']
. "'>Employees </a>)" . "</li>\n";
}
echo '</ul>';
$stmt = null;
}
catch (Exception $e) {
echo "Error";
echo $e->getMessage();
}
?>

Answer 1

The problem with the PHP script was that I was not specifying which tables to select from. This was harder than it should have been since the syntax above returned data in MySQL which was the database I was accessing, The correct PHP script that is working can be found below:
Working PHP:
<?php
try {
$sql = 'SELECT contact.FirstName,contact.LastName ';
$sql .= 'FROM adventureworks.employeedepartmenthistory ';
$sql .= 'JOIN adventureworks.employee USING(EmployeeID) ';
$sql .= 'JOIN adventureworks.contact USING(ContactID) ';
$sql .= 'WHERE DepartmentID = 2';
$stmt = $dbh->prepare($sql);
$stmt->execute();
echo $sql;
echo '<ul>';
foreach ($stmt->fetchAll() as $emps){
echo "<li>" . $emps["FirstName"] . $emps["LastName"] . "</li>";
}
echo '</ul>';
$stmt = null;
}
catch (Exception $e) {
echo "Error";
echo $e->getMessage();
}
?>
I was wrong that the tables would have been assumed through the JOIN in PHP as they are in MySQL.