我的SQL中的Select存在问题

时间:2016-05-24 14:13:50

标签: javascript php mysql select

记分牌未出现在右侧的高分数中。它没有显示任何东西。我为我的javascript蛇游戏制作了一个高分板。那么我的MYSQL代码有什么问题吗?

<html>

    <link href='style.css' type='text/css' rel='stylesheet'>

    <!-- Lets make a simple snake game -->
    <canvas id="canvas" width="450" height="450" style="border:1px solid #000000;"></canvas>

    <!-- Jquery -->
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" type="text/javascript">
    </script>
    <script src="snake.js" type="text/javascript"></script>


    <?php $score = "<script>document.write(score)</script>"?>

    <p id = 'demo'>Your Final Score: <script>score</script></p>

    <div id = 'right'>
       <div id = 'rightForm'>
          <form action="highscoreregistry.php" method="post">
             <input type="hidden" name="score" id="score">


          </form>
       </div>
       <div id = 'rightHighScore' style = 'size:9px;'>
        <?php
    $servername = "-";
    $username = "-";
    $password = "-";
    $dbname = "-";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT Name, Score, Date FROM snakehiscore";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    echo "<table><tr><th>ID</th><th>Name</th></tr>";
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<tr><td>".$row["Name"]."</td><td>".$row["Score"]." ".$row["Date"]."</td></tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}
$conn->close();
?>
   </div>
</div>


</html>

1 个答案:

答案 0 :(得分:-4)

  1. 在查询中添加分号!所以:

    $ sql =“SELECT名称,分数,日期来自snakehiscore;”;

  2. 2.而不是使用它:

    $result = $conn->query($sql);
    

    如果您安装了mysqli,请尝试此操作:

    $result = mysqli_query($conn, $sql);
    mysqli_num_rows($result) > 0