PHP 提取 MYSQL 值问题

Question

我有 2 个表：用户和事务

在我拥有的用户中：用户 ID、姓名、电子邮件

在交易中我有：id、idsender、idreceiver

我想制作一个显示所有交易的日志表，我希望它们显示为：交易 ID - SENDER'S NAME - RECEIVER'S NAME

我试过这样但它似乎不起作用，在“接收器”它没有显示任何东西..：

echo "<table>";
echo "<tr>";
echo "<th>Transaction ID</th>";
echo "<th>Sender</th>";
echo "<th>Receiver</th>";
echo "</tr>";

    $resultuser = mysqli_query($conn, "SELECT * FROM users"); 
    $rowuser=mysqli_fetch_array($resultuser);

    $resulttrans = mysqli_query($conn, "SELECT * FROM transactions"); 
    $rowtrans=mysqli_fetch_array($resulttrans);



    $ress=mysqli_query($conn, "SELECT * FROM transactions");

            while($row=mysqli_fetch_array($ress)){
                echo "<tr>";
                echo "<td>".$row['id']."</td>";
                echo "<td>"." same as receiver"."</td>";
            $receiver=mysqli_query($conn, "SELECT name FROM users WHERE userid='" . $rowtrans['idreceiver'] . "'");
            $receivername = mysqli_fetch_array($receiver);

                echo "<td>". $receivername ."</td>";
                echo "</tr>";
            }

Answer 1

不要使用两个查询。在一个查询中连接两个表：

$ress = mysqli_query("SELECT t.id, u1.name AS sendername, u2.name AS receivername
    FROM transactions AS t
    JOIN users AS u1 ON u1.userid = t.idsender
    JOIN users AS u2 ON u2.userid = t.idreceiver");
while ($row = mysqli_fetch_assoc($ress)) {
    echo "<tr>";
    echo "<td>".$row['id']."</td>";
    echo "<td>" . $row['sendername'] ."</td>";
    echo "<td>" . $row['receivername'] . "</td>";
    echo "</tr>";
}

Answer 2

您正在从 $receivername = mysqli_fetch_array($receiver); 中提取的数组中回显

你应该这样做echo "<td>". $receivername['name'] ."</td>";

注意： 从mysqli_fetch_array() 获取的数组可以使用编号$receivername[0] 或字符串$receivername['name'] 的索引来回显，这就是您在这里缺少的地方。