将行添加到html表

Question

我得到一个html表，从数据库获取信息并在水平和垂直方向上显示它。 我设法只显示用户名，但我需要添加全名和ID也。 这是我走到这一步：

查询：

$conn = mysqli_connect(DATABASE_HOST, DATABASE_USER, DATABASE_PASSWORD, DATABASE_NAME); 

// 1.Get data 
// data for final table 
// format is [username][projectNo] => [process1, process2, ..., processN] 
$result = []; 
// map project no to its title 
$projectNoToTitle = []; 
$sql = ' 
    SELECT uid, username, staff_id, longname 
    FROM `user` 
    ORDER BY username 
'; 
$query = mysqli_query($conn, $sql); 
// for each user 
while ($data = mysqli_fetch_assoc($query)) { 
    $sql2 = ' 
     SELECT a.* FROM 
     (  
      (
       -- select pairs project - leader 
       SELECT p.projectNo, p.title, CONCAT(pr.process, "(ld)") AS process  
       FROM project p 
       LEFT JOIN proc_leader pr ON p.projectNo = pr.projectNo 
        AND pr.proc_leader = "' . mysqli_real_escape_string($conn, $data['username']) . '" 
      ) 
      -- union all means we union result of queries, which have structure 
      -- and don\'t remove duplicates (it\'s faster than UNION and 
      -- more logical because in our sittuation it won\'t be any duplicates) 
      UNION ALL 
      (
       -- select pairs project - checker 
       SELECT p.projectNo, p.title, CONCAT(pch.process, "(chk)") AS process 
       FROM project p 
       LEFT JOIN proc_checker pch ON p.projectNo = pch.projectNo 
        AND pch.proc_checker = "' . mysqli_real_escape_string($conn, $data['username']) . '" 
      ) 
      UNION ALL 
      (
       -- select pairs project - checker 
       SELECT p.projectNo, p.title, CONCAT(pch.process, "(staff)") AS process 
       FROM project p 
       LEFT JOIN proc_staff pch ON p.projectNo = pch.projectNo 
        AND pch.proc_staff = "' . mysqli_real_escape_string($conn, $data['username']) . '" 
      ) 

     ) AS a 
     ORDER BY a.projectNo'; 
    $query2 = mysqli_query($conn, $sql2); 
    // for each project => process pair of user 
    while ($data2 = mysqli_fetch_assoc($query2)) { 
     $staff_id = $data['staff_id']; 
     $longname = $data['longname']; 
     $username = $data['username']; 
     $projectNo = $data2['projectNo']; 
     $projectTitle = $data2['title']; 
     $process = $data2['process']; 

     $projectNoToTitle[$projectNo] = $projectTitle; 

     if (!isset($result[$username])) { 
      $result[$username] = []; 
     } 
     if (!isset($result[$username][$projectNo])) { 
      $result[$username][$projectNo] = []; 
     } 
     if ($process) { 
      $result[$username][$projectNo][] = $process; 
     } 
    } 
} 

而且表：

  <table style="background-color:rgb(238, 238, 238)" id="dataTable4" class="tablesorter" class="tblD" border="0" cellpadding="0" cellspacing="1"> 
      <?php 

      // 2. Output table 
      // create table header 
      // it's columns should contain all projects 
      if ($result) { 
       $header = 
        '<th>Staff ID</th> 
        <th>Full Name</th> 
        <th>Username</th>' . 
        array_reduce(array_values($projectNoToTitle), function ($p, $n) { 
         return $p . '<th>Project ' . htmlspecialchars($n) . '</th>'; 
        }); 

       // output body 
       $body = ''; 
       foreach ($result as $username => $usernameData) { 
        $row = '<td>' . htmlspecialchars($username) . '</td>'; 
        foreach ($projectNoToTitle as $projectNo => $projectTitle) { 
         $r = isset($usernameData[$projectNo]) 
          ? implode(', ', $usernameData[$projectNo]) 
          : 'N/A'; 
         $row .= '<td>' . htmlspecialchars($r) . '</td>'; 
        } 
        $body .= "<tr>$row</tr>"; 
       } 

       echo "<thead>$header</thead><tbody>$body</tbody>"; 
      }// \2. Output table 
      ?> 
      </table> 

我如何添加$staff_id = $data['staff_id']; $longname = $data['longname'];表？

Answer 1

你可以在你的文章中更新数据库结构。