Php CRUD：在单个 while 循环中显示来自 2 个相关表的数据

Question

我制作了一个用于收集患者信息的 CRUD 应用程序。

数据库中有 2 个 table：users 和 medical_records。 "users" table 包含向 medical_records[=41 广告医疗记录的 医疗人员 =] table.

用户 id (uid) 是医疗记录中的外键 table。我想在 medical_records table.

中显示军医的全名

为此，我写了这个查询，首先：

$sql = "SELECT `first_name`, `last_name` FROM users WHERE id =`" . $row['uid'] . "`";

结果是想要的：

SELECT `first_name`, `last_name` FROM users WHERE id =`4;`

但是像这样回显全名 echo $row['first_name'] . $row['last_name']; 会导致错误：Undefined index: first_name in path\to\medical_records_all_table.php.

我试图从单个 while 循环中的两个 table 中提取数据的整个上下文是：

<table id="medical_records" class="table table-bordered table-striped" width="100%">
  <thead>
    <tr>
      <th>Nume complet</th>
      <th>Judet</th>
      <th>Data internarii</th>
      <th>Ora internarii</th>
      <th>Data operatiei</th>
      <th>Ora operatiei</th>
      <th>Fisa adaugata de</th>
      <th>Actiuni</th>
    </tr>
  </thead>
  <tbody>
    <?php
      $sql = "SELECT `mid`, `uid`, `nume_complet`, `judet`, `data_internarii_mamei`, `ora_internarii_mamei`, `data_operatiei`, `ora_operatiei` FROM medical_records";

      $result = mysqli_query($con, $sql);
      if (mysqli_num_rows($result) > 0) {
        while ($row = mysqli_fetch_assoc($result)) {
          $mid = $row['mid'];
          ?>
    <tr id="<?php echo $mid; ?>">
      <td>
        <a href="view_record.php?mid=<?php echo $mid; ?>">
        <?php echo ucwords($row['nume_complet']); ?>
        </a>
      </td>
      <td><?php echo $row['judet']; ?></td>
      <td><?php echo $row['data_internarii_mamei']; ?></td>
      <td><?php echo $row['ora_internarii_mamei']; ?></td>
      <td><?php echo $row['data_operatiei']; ?></td>
      <td><?php echo $row['ora_operatiei']; ?></td>
      <td>
        <?php
          echo $sql = "SELECT `first_name`, `last_name` FROM users WHERE id =`" . $row['uid'] . ";`";
          echo $row['first_name'] . $row['last_name'];
        ?>
      </td>
      <td class="actions">
        <ul class="list-inline text-center">
          <li>
            <a title="Vezi fisa" href="view_record.php?mid=<?php echo $mid; ?>"><span class="glyphicon glyphicon-eye-open"></span></a>
          </li>
          <li>
            <a title="Editeaza" href="edit_record.php?mid=<?php echo $mid; ?>"><span class="glyphicon glyphicon-edit"></span></a>
          </li>
          <li>
            <a title="Sterge" class="delete-icn" href="#" data-mid="<?php echo $mid; ?>"><span class="glyphicon glyphicon-trash"></span></a>
          </li>
        </ul>
      </td>
    </tr>
    <?php } mysqli_free_result($result); } ?>
  </tbody>
</table>

我错过了什么？

Answer 1

尝试

$sql = "SELECT `first_name`, `last_name` FROM users WHERE id =`" . $row['uid'] . ";`";
$result = mysqli_query($con, $sql);
while ($row = mysqli_fetch_assoc($result)) {
 echo $row['first_name'] . $row['last_name'];
}

而不是

echo $sql = "SELECT `first_name`, `last_name` FROM users WHERE id =`" . $row['uid'] . ";`";
 echo $row['first_name'] . $row['last_name'];

最好的方法是在第一个 sql.

中加入用户和 medical_records table，如下所示

$sql = "SELECT m.mid, m.uid, m.nume_complet, m.judet, m.data_internarii_mamei, m.ora_internarii_mamei, m.data_operatiei, m.ora_operatiei, u.first_name, u.last_name FROM medical_records m join users u on m.uid = u.id";

以下不需要

echo $sql = "SELECT `first_name`, `last_name` FROM users WHERE id =`" . $row['uid'] . ";`";