Warning: mysqli_close() expects parameter 1 to be

问题:

I have this PHP code on a WAMP server with MySQL and this error appears in a dialog box when I visit the site. How can I fix this?

<?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);
?>

回答1:

You don't need this line:

mysqli_close($result);

You only need to close the connection.