Javascript jquery使用php和mysql获取语法以进行输入验证

这件事让我很沮丧。 保持简单。 我有一个带有输入和p元素的网页

<input type=text name="findSong" id="findSong">
<p id = " songCheck">Search</p>

试试这个：

    <?php
if(isset($_POST['findSong'])){
    $servername = "localhost";
    $database = "karaoke";
    $username = "root";
    $password = "";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $database);

    // Check connection

    if (!$conn) {

        die("Connection failed: " . mysqli_connect_error());

    }
    $song = $_POST["findSong"]);
    $sql = "SELECT * FROM songlist WHERE SongNum = '$song'"; 
    if (mysqli_query($conn, $sql)) {
          echo "Match Found"
    } else {
          echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

    ?>

---

我们需要查看jquery code.oops抱歉的编辑问题！
“$song.”“；
中有一个额外的括号，说明了您所做的工作，以及如何解决问题，从而改进了答案。感谢您的回答，这更多是我的jquery，而不是按键上的任何操作
$("input").keypress(function(){
    $.get("songCheck.php", function(data, status){
        document.getElementById('songCheck').innerHTML = data;
    });
});

    <?php
if(isset($_POST['findSong'])){
    $servername = "localhost";
    $database = "karaoke";
    $username = "root";
    $password = "";

    // Create connection
    $conn = mysqli_connect($servername, $username, $password, $database);

    // Check connection

    if (!$conn) {

        die("Connection failed: " . mysqli_connect_error());

    }
    $song = $_POST["findSong"]);
    $sql = "SELECT * FROM songlist WHERE SongNum = '$song'"; 
    if (mysqli_query($conn, $sql)) {
          echo "Match Found"
    } else {
          echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
}

    ?>