Php CRUD：在单个while循环中显示来自2个相关表的数据

Question

我做了一个CRUD应用程序来收集有关患者的信息。

数据库中有2个表格：用户和medical_records。 “用户”表中包含医疗个人即广告病历，换句话说，数据为medical_records表。

用户标识（uid）是医疗记录表中的外键。我想在medical_records表中显示医师的全名。

为此，我写了这个查询，首先：

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

并将结果所需的一个：

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

但呼应的全名像这样echo $row['first_name'] . $row['last_name'];会导致错误：Undefined index: first_name in path\to\medical_records_all_table.php。

在那里我试图提取里面一个从两个表中的数据，而循环是整个背景：

<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> 

我缺少什么？

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']; 

最好的办法是如下加入用户和medical_records表中的第一个SQL。

$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"; 

和下方不需要

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