我有一个MySQL的WAMP的服务器上这个PHP代码，当我访问该网站会出现在一个对话框这个错误。 我怎样才能解决这个问题？
<?php
// Create connection
$con = mysqli_connect("localhost","userdb","userdb","apptestdb");
// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// This SQL statement selects ALL from the table 'Locations'
$sql = "SELECT * FROM Locations";
// Check if there are results
if ($result = mysqli_query($con, $sql))
{
// If so, then create a results array and a temporary one
// to hold the data
$resultArray = array();
$tempArray = array();
// Loop through each row in the result set
while($row = $result->fetch_object())
{
// Add each row into our results array
$tempArray = $row;
array_push($resultArray, $tempArray);
}
// Finally, encode the array to JSON and output the results
echo json_encode($resultArray);
}
// Close connections
mysqli_close($result);
mysqli_close($con);
?>
