MySQL 2行返回，只有1顯示內容

Question

因此，我有一個問題與MySQL返回2行，但只有1顯示在變量。 不要知道爲什麼它沒有返回兩個

 $sql->bind_result($c_id, $c_location, $c_type) or die($mysqli_load->error); 
     while($row = $sql->fetch()){ 
     $mysqli_load2 = new mysqli(HOST, USER, PASS, DB); 
     $query = "SELECT `badge` FROM `responders` WHERE `cid` = ?"; 
     $sql2 = $mysqli_load2->prepare($query) or die($mysqli_load2->error); 
     $sql2->bind_param('i', $c_id); 
     $sql2->execute() or die($mysqli_load2->error); 
     $sql2->store_result(); 
     $rows = $sql2->num_rows; 
     $sql2->bind_result($units); 
     $sql2->fetch(); 
     $sql2->close(); 
     $mysqli_load2->close(); 
      echo ' 
      <tr> 
      <td align="justify"><a href="viewcall.php?cid=' . $c_id .'"><u><abbr title="View Call">'.$c_id.'</abbr></u></td> 
      <td align="justify">' . $c_location .'</td> 
      <td align="justify">' . $c_type .'</td> 
      <td align="justify">Rows: ' . $rows . ' ' . $units . '</td></tr>'; 
      } 

Answer 1

您需要循環使用fetch（）方法：對於任何人誰也有類似的問題

while ($sql2->fetch()) { 
    echo $units; 
} 
$sql2->close(); 

Answer 2

因此，這裏是新更新的代碼這個：

 $sql->bind_result($c_id, $c_location, $c_type) or die($mysqli_load->error); 
     while($row = $sql->fetch()){ 
      $mysqli_load2 = new mysqli(HOST, USER, PASS, DB); 
      $query = "SELECT `badge` FROM `responders` WHERE `cid` = ?"; 
      $sql2 = $mysqli_load2->prepare($query) or die($mysqli_load2->error); 
      $sql2->bind_param('i', $c_id); 
      $sql2->execute() or die($mysqli_load2->error); 
      $sql2->bind_result($units); 
        echo ' 
      <tr> 
      <td align="justify"><a href="viewcall.php?cid=' . $c_id .'"><u><abbr title="View Call">'.$c_id.'</abbr></u></td> 
      <td align="justify">' . $c_location .'</td> 
      <td align="justify">' . $c_type .'</td> 
      <td align="justify">'; 
      while($sql2->fetch()) 
      { 
      echo 
      '#'.$units.' '; } echo '</td></tr>'; 
      } 
      $sql2->close(); 
      $mysqli_load2->close(); 
      $sql->close(); 
      $mysqli_load->close();