Добавление строк в HTML-таблицу с динамическими столбцами

Question

Я использую ниже код (при условии, как решение предыдущего вопроса на Display mySQL records as HTML table columns), чтобы сгенерировать HTML-таблицу, как:

На основе этих столы:

неделя имена в таблице «недели» и каждую неделю запись также содержит количество сессий на этой неделе:

+---------+-----------+----------+-----------+ 
| week_pk | week_name | sessions | cohort_fk | 
+---------+-----------+----------+-----------+ 
|  1 | Week 1 |  3 |   1 | 
|  2 | Week 2 |  2 |   1 | 
|  3 | Week 3 |  1 |   1 | 
+---------+-----------+----------+-----------+ 

+-----------+-------------+-------------+-------------+ 
| cohort_pk | cohort_name | cohort_code | cohort_year | 
+-----------+-------------+-------------+-------------+ 
|   1 | Some name | MICR8976 |  2014 | 
+-----------+-------------+-------------+-------------+ 

Теперь я хочу расширить это, чтобы показать посетителям в виде дополнительных строк таблицы в строке сеанса и указать, на какой сессии они присутствовали. В таблице посещаемость:

+---------------+-------------+---------+-----------+---------+---------+ 
| attendance_pk | given_names | surname | cohort_fk | week_fk | session | 
+---------------+-------------+---------+-----------+---------+---------+ 
|    1 | Bill  | Smith |  1 |  2 |  2 | 
|    2 | Fred  | Jones |  1 |  1 |  1 | 
+---------------+-------------+---------+-----------+---------+---------+ 

Результирующий HTML таблица будет как:

В любом случае помощь по модификации кода ниже, чтобы получить результаты согласно изображению выше ценится.

$cohort = '1'; 
$year = '2014'; 

    $query = "SELECT * FROM cohort, week, attendance 
    WHERE week.cohort_fk = cohort.cohort_pk 
    AND attendance.week_fk = week.week_pk 
    AND attendance.cohort_fk = cohort.cohort_pk 
    AND cohort.cohort_year = '$year' 
    AND cohort.cohort_pk = '$cohort'"; 

    $result = mysql_query($query, $connection) or die(mysql_error()); 

    echo "<table border='1'>"; 
    echo "<tr><td>Name</td>"; 
    $second_row = "<tr><td>Session</td>"; 
    while($row = mysql_fetch_assoc($result)){ 
     $weekname = $row["week_name"]; 
     $n_session = $row["sessions"]; 
     echo "<td colspan='$n_session'>$weekname</td>"; 
     for($i=1; $i<=$n_session; $i++){ 
      $second_row .= "<td>S$i</td>"; 
     } 
    } 
    echo "</tr>"; 
    echo "$second_row</tr>"; 
    echo "</table>"; 
    ?> 

Answer 1

Следующий код может быть самым простым &, ближайшим к тому, что вы хотите.

• mysqli_* функции используются.
• Включены два запроса.
• Первый запрос извлекается только с cohort и week.
• В первой while петля $weeksession массив создан. Он содержит номер столбца для данного номер недели и номер сеанса.

---

$year = 2014; 
$cohort = 1; 

$query = "SELECT * FROM cohort, week 
WHERE week.cohort_fk = cohort.cohort_pk 
AND cohort.cohort_year = '$year' 
AND cohort.cohort_pk = '$cohort' 
ORDER BY week.week_pk"; 

$dblink = mysqli_connect("localhost", "root", "", "test"); 
$result = mysqli_query($dblink, $query); 

echo "<table border='1'>"; 
echo "<tr><td>Name</td>"; 
$second_row = "<tr><td>Session</td>"; 
$totalcolumn = 1;        
while($row = mysqli_fetch_assoc($result)){ 
    $weekname = $row["week_name"]; 
    $n_session = $row["sessions"]; 
    $weekpk  = $row["week_pk"];    
    $totalcolumn += $n_session;     
    echo "<td colspan='$n_session'>$weekname</td>"; 
    for($i=1; $i<=$n_session; $i++){ 
     $second_row .= "<td>S$i</td>"; 
     $weeksession[$weekpk][$i] = $totalcolumn - $n_session + $i; 
    } 
}//end while 
echo "</tr>"; 
echo $second_row . "</tr>"; 


$query = "SELECT * FROM cohort, week, attendance 
WHERE week.cohort_fk = cohort.cohort_pk 
AND attendance.week_fk = week.week_pk 
AND attendance.cohort_fk = cohort.cohort_pk 
AND cohort.cohort_year = '$year' 
AND cohort.cohort_pk = '$cohort' 
ORDER BY attendance.attendance_pk"; 
$result = mysqli_query($dblink, $query); 
while($row = mysqli_fetch_assoc($result)){ 
    $name = $row["given_names"] . " " . $row["surname"]; 
    $weekpk  = $row["week_pk"]; 
    $sno  = $row["session"]; 
    echo "<tr><td>$name</td>"; 
    for($i=2; $i<=$totalcolumn; $i++){  
     if($weeksession[$weekpk][$sno] == $i) 
      echo "<td>X</td>"; 
     else 
      echo "<td>-</td>";    
    }          
    echo "</tr>"; 
}//end while 
echo "</table>"; 

---

• Если человек посещает 2 или более сессий, то он будет показан в нескольких строках

Answer 2

может быть, это помогает.

<?php 

$year = "2014"; 
$cohort = 1; 

$query = "SELECT * FROM cohort, week WHERE week.cohort_fk = cohort.cohort_pk AND cohort.cohort_year = '$year' AND cohort.cohort_pk = '$cohort'"; 
$result = mysql_query($query, $connection) or die(mysql_error()); 

$rows = array(); 
$weeks = array(); 
$sessions = array(); 
while ($rows[] = $row = mysql_fetch_array($result, MYSQL_ASSOC)) { 
    $weeks[] = "<td colspan='" . (int) $row['sessions'] . "'>" . $row['week_name'] . "</td>"; 
    for ($i = 1; $i <= (int) $row['sessions']; $i++) { 
     $sessions[] = "<td>S" . $i . "</td>"; 
    } 
} 

$attendance = array(); 
$query = "SELECT * FROM cohort, week, attendance 
WHERE week.cohort_fk = cohort.cohort_pk 
AND attendance.week_fk = week.week_pk 
AND attendance.cohort_fk = cohort.cohort_pk 
AND cohort.cohort_year = '$year' 
AND cohort.cohort_pk = '$cohort' 
ORDER BY attendance.attendance_pk"; 
$result = mysql_query($query, $connection) or die(mysql_error()); 
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { 
    $attendance[] = "<tr><th>" . $row['given_names'] . " " . $row['surname'] . "</th>"; 
    foreach ($rows as $week) { 
     for ($i = 1; $i <= (int) $week['sessions']; $i++) { 
      if ($row['week_fk'] == $week['week_pk'] && $row['session'] == $i) { 
       $attendance[] = "<td>X</td>"; 
      } else { 
       $attendance[] = "<td></td>"; 
      } 
     } 
    } 
    $attendance[] = "</tr>"; 
}; 

$table = array(
    "<table width = 50% border = '1' cellspacing = '2' cellpadding = '0'>", 
    "<tr><th>Name</th>", join('', $weeks), "</tr>", 
    "<tr><th>Sessions</th>", join('', $sessions), "</tr>", 
    join('', $attendance), 
    "</table>" 
); 

echo join("", $table);