PHP 结果没有获取 MySQL 数据库

Question

我不懂 PHP 和 SQL。我们只是在学期末才勉强把它刮掉，这让我很沮丧。我试图让我的结果页面显示正确的信息，但对于我来说，它不会抓取任何东西。我显然有问题，想知道是否可以得到一些帮助。

初始页

(这里是基本网页的正常顶部)

    <form id="ClubForm" action="ClubMembersResults.php" method="get">

    <?php            
        require_once ('dbtest.php');
        $query= "SELECT * FROM tblMembers ORDER BY LastName, FirstName, MiddleName;";
        $r = mysqli_query($dbc, $query);
        if (mysqli_num_rows($r) > 0) {
            echo '<select id="memberid" name="memberid">';
            while ($row = mysqli_fetch_array($r)) {
                echo '<option value="'.$row['LastName'].'">'
                        .$row['LastName'].", ".$row['FirstName']." ".$row['MiddleName']. '</option>';
            }
            echo '</select>';
        } else {
            echo "<p>No Members found!</p>";
        }
    ?>
    <input type="submit" name="go" id="go" value="Go" />
    </form>
    <div id="results"></div>
</body>

结果页面当前写为:

<?php
        $memid = 0;
        $memid = (int)$_GET['memberid'];
        if ($memid > 0) {

            require_once ('dbtest.php');
            $query = "SELECT * FROM tblMembers WHERE MemID = $memid;";
            $r = mysqli_query($dbc, $query);`enter code here`
            if (mysqli_num_rows($r) > 0) {
            $row = mysqli_fetch_array($r);
            echo "<p>Member ID: ".$row['MemID']."<br>";
            echo "Member Name: ".$row['LastName'].", ".$row['FirstName']." ".$row['MiddleName']."<br>";
            echo "Member Joined: ".$row['MemDt']."<br>";
            echo "Member Status: ".$row['Status']."<br></p>";
            }else {
                echo "<p>Member not on file.</p>";
            }
            //table for inverntory
            echo "<table border='1'>";
            echo "<caption>Transaction History</caption>";
            echo "<tr>";
            echo "<th>Purchase Dt</th>";
            echo "<th>Trans Cd</th>";
            echo "<th>Trans Desc</th>";
            echo "<th>Trans Type</th>";
            echo "<th>Amount</th>";                
            echo "</tr>";
            $query2 = "SELECT p.Memid, p.PurchaseDt, p.TransCd, c.TransDesc, p.TransType, p.Amount
                       FROM tblpurchases p, tblcodes c
                       WHERE p.TransCd = c.TransCd AND p.MemId = 'member id'
                       ORDER BY p.MemId, p.PurchaseDt, p.TransCd
                       ";
            $r2 = mysqli_query($dbc, $query2);
            while ($row = mysqli_fetch_array($r2)) {
                echo "<tr>";
                echo "<td>".$row['PurchaseDt']."</td>;";
                echo "<td>".$row['TransCd']."</td>";
                echo "<td>".$row['TransDesc']."</td>";
                echo "<td>".$row['TransType']."</td>";
                echo "<td>".$row['Amount']."</td>";
                echo "</tr>";
            }
            echo "</table>";                
        } else {
            echo '<p>No Member ID from form.</p>';
        }
?>

结果页面应该显示带有 TH 和 TR/TD 区域信息的表格。这两个区域都来自单独的 SQL 表，并且 tblmembers 和 tblpurchases 之间唯一相似的字段是 MemID。

Answer 1

您需要在 sql 请求中使用 join 来显示成员(member)的购买情况。

  SELECT m.Memid, p.PurchaseDt, p.TransCd, 

                   FROM tblpurchases p join 
  tblmembers m on p.MemId=m.MemId

这是一个连接的例子