MYSQL/PHP回送外键参数的平均值

我试图弄清楚如何让两条SQL语句背靠背，这将创建一个表，其中包含第一条SQL语句中的信息，最后一个单元格是该表中一列的平均值

我可以显示制作表格所需的信息。没问题。是这样的：

$sql = "SELECT golfer_name, golfer_handicap, golfer_ghin FROM golfers WHERE trip_name_table_ID = '$userid'";
$result = $mysqli->query($sql);

if ($result->num_rows > 0) {
    while($rowitem = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td style='text-align: left'>" . $rowitem['golfer_name'] . "</td>";
        echo "<td>" . $rowitem['golfer_handicap'] . "</td>";
        echo "<td>" . $rowitem['golfer_ghin'] . "</td>";
        echo "</tr>";

---

但就我的绝对生命而言，我不知道如何

回显
那一条数据。
你只会像第一次查询那样去做；保持变量不同，以免干扰循环

<?php
$sql = "SELECT golfer_name, golfer_handicap, golfer_ghin FROM golfers WHERE trip_name_table_ID = '$userid'";
$result = $mysqli->query($sql);

$sql2 = "SELECT CAST(AVG(golfer_handicap) AS DECIMAL (3,1)) AS handicap FROM golfers WHERE trip_name_table_ID = '$userid'";
$result2 = $mysqli->query($sql2);
$total = mysqli_fetch_array($result2);

if ($result->num_rows > 0) {
    while($rowitem = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td style='text-align: left'>" . $rowitem['golfer_name'] . "</td>";
        echo "<td>" . $rowitem['golfer_handicap'] . "</td>";
        echo "<td>" . $rowitem['golfer_ghin'] . "</td>";
        echo "</tr>";

解决方案：感谢@Dawson\u Irvine，请看他们是如何解决我的问题的，而不是通读评论。享受吧

<?php
$sql = "SELECT golfer_name, golfer_handicap, golfer_ghin FROM golfers WHERE trip_name_table_ID = '$userid'";
$result = $mysqli->query($sql);

$sql2 = "SELECT CAST(AVG(golfer_handicap) AS DECIMAL (3,1)) AS handicap FROM golfers WHERE trip_name_table_ID = '$userid'";
$result2 = $mysqli->query($sql2);
$total = mysqli_fetch_array($result2);

我用“$average”代替“$total”，并在我希望它被调用的地方输入了这个代码：
echo“average disability:”.$average['golfer_disability']。"";并且它使我的表消失，并且没有调用所需的值。我已经稍微更新了代码。我确实犯了一个错误，这会妨碍平均数的计算。在
$result2=mysqli->query（$sql2）上我输入了一些错误，但现在应该更正。好吧，现在这至少让表格恢复了，显示了“平均障碍：”文本，但平均数据仍然没有显示
echo
code与上面的注释相同。在$sql2查询中，我已使查询返回数组中的平均值作为“disability”键。尝试
$average['disability']
或将
更改为
高尔夫球手(高尔夫障碍)
宾果游戏。我认为这很有效！谢谢介意快速解释一下AS是干什么的吗？因为它显然没有在我的表中创建一个新列，因为它不是INSERT。。所以我猜这是一个“假的”专栏？
<?php
$sql = "SELECT golfer_name, golfer_handicap, golfer_ghin FROM golfers WHERE trip_name_table_ID = '$userid'";
$result = $mysqli->query($sql);

$sql2 = "SELECT CAST(AVG(golfer_handicap) AS DECIMAL (3,1)) AS handicap FROM golfers WHERE trip_name_table_ID = '$userid'";
$result2 = $mysqli->query($sql2);
$total = mysqli_fetch_array($result2);

if ($result->num_rows > 0) {
    while($rowitem = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td style='text-align: left'>" . $rowitem['golfer_name'] . "</td>";
        echo "<td>" . $rowitem['golfer_handicap'] . "</td>";
        echo "<td>" . $rowitem['golfer_ghin'] . "</td>";
        echo "</tr>";

<?php
$sql = "SELECT golfer_name, golfer_handicap, golfer_ghin FROM golfers WHERE trip_name_table_ID = '$userid'";
$result = $mysqli->query($sql);

$sql2 = "SELECT CAST(AVG(golfer_handicap) AS DECIMAL (3,1)) AS handicap FROM golfers WHERE trip_name_table_ID = '$userid'";
$result2 = $mysqli->query($sql2);
$total = mysqli_fetch_array($result2);

echo "<td colspan = '2'> Average Handicap: " ."<b>". $average['handicap'] ."</b></td>";