Php 如何显示MySQL表中的平均值

我有一个MySQL数据库，其中包含各种视频游戏和用户帐户的详细信息。用户可以查看游戏，并将分数放入查看表中。 我试图检索每场比赛的平均分数，并将其显示在网页上。我在PHP中的SQL语句如下所示

$sqlOverall = "SELECT CAST(AVG('scoreOverall') AS DECIMAL('10,1')) 
               FROM 'ratings' 
               WHERE gameID = '$id'";
$scoreOverall = mysqli_query($conn, $sqlOverall);

---

但当我尝试回显$scoreOverall时，它返回空白。

尝试下面的查询

$selquery="SELECT CAST(AVG(scoreOverall) AS DECIMAL(10,1)) FROM ratings WHERE gameID='$id'";
$result= mysqli_query($conn, $selquery);

试试上面的代码。

---

希望这会有所帮助。

效果很好。希望能对你有所帮助

$sqlOverall="SELECT CAST(AVG(scoreOverall) AS DECIMAL(10,1)) FROM ratings WHERE gameID=$id";
$scoreOverall = mysqli_query($conn, $sqlOverall);
foreach ($scoreOverall as $value) { 
    $scoreOverall = $value['CAST(AVG(queue_status) AS DECIMAL(10,1))']; 
    echo $scoreOverall;
}

---

当我尝试回显$scoreOverall时，$result变量包含一个对象（类型为mysqli_result），您可以从中获取需要输出的scoreOverall。echo$result->fetch_object（）->scoreOverall；返回错误“可捕获致命错误：在第77行中，类mysqli_result的对象无法转换为字符串”。在第77行，我回显$id，它在我尝试此SQL语句之前回显

$sqlOverall="SELECT CAST(AVG(scoreOverall) AS DECIMAL(10,1)) FROM ratings WHERE gameID=$id";
$scoreOverall = mysqli_query($conn, $sqlOverall);
foreach ($scoreOverall as $value) { 
    $scoreOverall = $value['CAST(AVG(queue_status) AS DECIMAL(10,1))']; 
    echo $scoreOverall;
}