Question

我有一个从数据库中获取数据的html表。

这是我的SQL查询：

// 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)) {

    $sql2 = '
        SELECT a.* FROM
        (
            (
              -- select pairs project - leader
              SELECT p.projectNo, p.title, CONCAT(upr.process, (upr.role) ) AS process
              FROM project p
                LEFT JOIN user_project upr ON p.projectNo = upr.projectNo
                  AND upr.username = "' . mysqli_real_escape_string($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;
        }
    }
}

然后我想在表格中水平和垂直打印数据：

       <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) {
        $row = '<td>' . htmlspecialchars($longname) . '</td>' . '<td>' . htmlspecialchars($staff_id) . '</td>' . '<td>' . htmlspecialchars($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
?>

我可以打印用户名，但staff_id和longname出现问题。这是我现在的输出。

系统从表用户名中获取姓氏，并为列表中的每个用户名打印

Answer 1

问题是您没有将$longname和$staff_id放入$results数组中。打印表时，只需使用这些变量，其中包含数据库中最后一个用户的值。

将处理数据库结果的循环更改为：

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] = [ 'longname' => $longname, 'staff_id' => $staff_id, 'projects' => []];
    }
    if (!isset($result[$username]['projects'][$projectNo])) {
        $result[$username]['projects'][$projectNo] = [];
    }
    if ($process) {
        $result[$username]['projects'][$projectNo][] = $process;
    }
}

然后构建表的代码应为：

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

将数据添加到html表

1 个答案: