求助：关于接收Ajax调用的PHP？

Question

我遇到了一个我找不到答案的问题，我很少在这里问问题，但我很困惑。如有任何协助，将不胜感激。

这是接收Ajax调用的PHP：

<?php
    session_start();
    $_SESSION["my_data"] = $_POST['action'];

    $DB_HOSTNAME = 'localhost';
    $DB_USERNAME = 'username';
    $DB_PASSWORD = 'password';

    $link2 = mysqli_connect($DB_HOSTNAME,$DB_USERNAME,$DB_PASSWORD) or die('Unable to establish a DB1 connection');
    mysqli_select_db($link2, '$DB_USERNAME');

    $orderQuery = mysqli_query($link2, "SELECT * FROM table WHERE id='".$_SESSION['my_data']."'");
    $orderQuery = mysqli_fetch_assoc($orderQuery);

    $orderInfo = "
        <table class='table table-striped'>
        <tbody>
            <tr>
                <td>#: </td>
                <td>". $_SESSION['my_data'] ."</td>
            </tr>
            <tr>
                <td> Full name: </td>
                <td>". $orderQuery['firstname'] . " " . $orderQuery['lastname'] ."</td>
            </tr>
            <tr>
                <td> Address: </td>
                <td> ". $orderQuery['shipping_address_1'] ."<br> ". $orderQuery['shipping_city'] . " " . $orderQuery['shipping_zone'] . " " . $orderQuery['shipping_postcode'] ." </td>
            </tr>
            <tr>
                <td> Card Expiry Date Month: </td>
                <td> 08 </td>
            </tr>        
        </tbody>
    </table><br>
    ";

    echo $orderInfo/* . $_POST['action']*/;   ?>

这是进行调用的脚本：

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script>

    var myWindow;
    function myFunction() {
    myWindow = window.open('', '_blank');
    myWindow.document.write("<link rel='stylesheet' type='text/css' href='stylesheet.css'>");

    var orderNum;
    orderNum = document.getElementsByClassName('summary_value')[0].innerHTML;
    orderNum = orderNum.replace("#", "");

    $.ajax( { type : 'POST',
      data : {'action':orderNum},
      url  : 'process.php',
      success: function ( data ) {
        myWindow.document.write( data );
      },
      error: function ( xhr ) {
        alert( "error" );
      }
    });

    myWindow.document.write("<br>");
    myWindow.document.write(document.getElementById("payInfor").innerHTML);
    }

</script>
<button onclick='myFunction()' class="btn btn-default">Print Pay Info</button>

这可能是一个简单的问题，但我没能理解。

我知道Ajax正在工作，因为它显示了一些信息（$ _SESSION ['my_data']），所以我认为这与我的SQL语句有关，但语法看起来还是正确的。

Answer 1

mysqli_select_db() expects parameter one to be the connection and parameter two to be the database name. In your case you are passing '$DB_USERNAME'. So your code will look to connect to database named '$DB_USERNAME' because of the single quotes. Change it to the database name instead. Either $DB_NAME or 'database_name' should work. While in development mode, try enabling error_reporting(E_ALL) to catch errors like this.