PHP mysqli_query将不会执行

下面的代码 – 我在那里添加了一些ECHO,以确保一切顺利.在$result = mysqli_query之前,似乎我们都很好.

<?php

//header('Content-type:application/json');

    // Connect to db
    $con = mysqli_connect("HOSTNAME","USER","PASSWORD","DATABASE");

    echo "Database: ";
    // Check connection
    if ($con->connect_error) {
        die("Connection failed: " . $con->connect_error);
    } 
    echo "Connected successfully";
    echo "<br><br>";

    // Get value from url
    $bid = $_GET['bid'];
    echo "BID: ";
    echo $bid;
    echo "<br><br>";

    // Define Query
    $sql = "SELECT id,bandname,members,bio,songlist FROM bands WHERE id='$bid'";
    echo "SQL Query: ";
    echo $sql;
    echo "<br><br>";

    // Run Query
    $result = mysqli_query($con,$sql);
    echo "Result: ";
    print_r($result);
    echo "<br><br>";

    // Put Query result into array
    $result_array = mysqli_fetch_array($result,MYSQLI_ASSOC);
    echo "Result Array: ";
    print_r($result_array);
    echo "<br><br>";

    // Encode array as JSON and output
    echo "JSON: ";
    echo json_encode($result_array);

?>

我输入的网址是http://bandsly.com/api.php?bid=1,这是我在浏览器中输出的输出…

Database: Connected successfully

BID: 1

SQL Query: SELECT id,songlist FROM bands WHERE id='1'

Result: 

Result Array: 

JSON: null

当我手动运行查询SELECT id,songlist FROM band WHERE id =’1’在数据库(PHPmyadmin)中,它工作正常,我得到1行返回正确的值.

手动db查询结果：

(http://i.stack.imgur.com/d3ZPJ.png)

非常感激任何的帮助！！

*****编辑*****

好吧,我想我发现了这个问题……我的“成功连接”没有正确写入,并且总是回来看起来很好.它看起来像修复后,我有数据库连接问题.

我要去查找数据库连接设置并尝试修复它.

感谢你的帮助！