Question

我有以下代码：

<?php
  include_once '../includes/db.inc.php';
  $sql = "SELECT * FROM clients ORDER BY nif_id ASC;";
  $result = mysqli_query($conn, $sql);
  $resultCheck = mysqli_num_rows($result);

  if ($resultCheck > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
      $first = $row["prm_nome"];
      $last = $row["apelido"];
      $phone = $row['nmr_tlm'];
      $email = $row['mail'];
      $nif = $row['nif_id'];
      $flight = "SELECT flight_id FROM flights INNER JOIN clients ON flights.nif_id=clients.nif_id";

      echo '<tr>';
      echo '<td><a href="detail.php?id='. $nif . '">'.$nif.'</a></td>';
      echo '<td>'.$first.'</td>';
      echo '<td>'.$last.'</td>';
      echo '<td>'.$phone.'</td>';
      echo '<td>'.$email.'</td>';
      echo '<td><a href="../flights/detail.php?id='. $flight . '">'.$flight.'</a></td>';
      echo '</tr>';
    }
  }
 ?>

我需要回显SELECT flight_id FROM flights INNER JOIN clients ON flights.nif_id=clients.nif_id查询的结果。但是当我保存文件时，页面上显示的是该查询的链接，而不是结果。

是否应该使用第一个$sql =下的查询来开始新的$sql =？还是有其他方法？

我尝试了UNION和SELECT *, flight_id FROM flights INNER JOIN clients ON flights.nif_id=clients.nif_id FROM clients ORDER BY nif_id ASC;，但后来得到了mysqli_num_rows() expects parameter 1 to be mysqli_result, bool given。

Answer 1

您不需要两个查询。只需使用加入的查询即可。

class SubjectDetailView(generic.DetailView, SingleObjectMixin):
    model           = Programmesearch
    template_name   = 'mnsdirectory/subject_detail.html'

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)
        # subjectslug = self.object.slug # Not necessary since It's a DetailView
        studyLevel = StudyLevel.objects.filter(
                            study_level=self.kwargs.get('studylevel', None))

在表上回显两个SQL查询结果

1 个答案: