动态HTML表 - 更新单元格行
我有一个从数据库动态生成的HTML表。在这个问题Adding rows to a HTML table with dynamic columns中可以看到实现这一目标的解决方案。
这样可以正常工作，除了我想在同一行中指出一个人每周参加的所有会话 - 使用下面的代码，额外的会话出席成为附加到HTML表的另一行。所以我想要的是：
数据库表就像（＆＃39;周＆＃39;，＆＃39;同期＆＃39;＆＃39;出勤＆＃39;表）
+---------+-----------+----------+-----------+
| 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 |
+-----------+-------------+-------------+-------------+
+---------------+-----------+-------------+---------+-----------+---------+---------+
| attendance_pk | person_id | given_names | surname | cohort_fk | week_fk | session |
+---------------+-----------+-------------+---------+-----------+---------+---------+
| 1 | 123456 | Bill | Smith | 1 | 1 | 2 |
| 2 | 123456 | Bill | Smith | 1 | 2 | 2 |
| 3 | 753354 | Fred | Jones | 1 | 1 | 1 |
| 4 | 753354 | Fred | Jones | 1 | 2 | 1 |
| 5 | 753354 | Fred | Jones | 1 | 3 | 1 |
+---------------+-----------+-------------+---------+-----------+---------+---------+
我正在使用的代码：
$cohort = $_POST['cohort'];
$year = $_POST['year'];
$query = "SELECT * FROM cohort, week
WHERE week.cohort_fk = cohort.cohort_pk
AND cohort.cohort_year = '$year'
AND cohort.cohort_pk = '$cohort'
ORDER BY week.week_pk";
$result = mysql_query($query, $connection) or die(mysql_error());
echo "<table width='100%' cellpadding='4' cellspacing='0' class='attendance_table'>";
echo "<tr><td class='theadings'></td>";
$second_row = "<tr><td class='theadings'></td>";
$totalcolumn = 1;
while( $row = mysql_fetch_assoc($result) ){
$weekname = $row["week_name"];
$n_session = $row["sessions"];
$weekpk = $row["week_pk"];
$totalcolumn += $n_session;
echo "<td class='theadings' colspan='$n_session'>$weekname</td>";
for($i=1; $i<=$n_session; $i++){
$second_row .= "<td class='theadings_lab'>Lab $i</td>";
$weeksession[$weekpk][$i] = $totalcolumn - $n_session + $i;
}
}
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";
$result = mysql_query($query, $connection) or die(mysql_error());
while( $row = mysql_fetch_assoc($result) ){
$name = $row["given_names"] . " " . $row["surname"];
$weekpk = $row["week_pk"];
$sno = $row["session"];
echo "<tr><td class='tborder_person_left'>$name</td>";
for($i=2; $i<=$totalcolumn; $i++){
if( $weeksession[$weekpk][$sno] == $i )
echo "<td class='tborder_person_attended'>✔</td>";
else
echo "<td class='tborder_person'>-</td>";
}
echo "</tr>";
}//end while
echo "</table>";
下面的@Kickstart是表格与您的代码相似的示例。你可以看到例如Melody Chew和Kit Yeng Melody Chew（同一个人）有两个单独的行。唯一标识符必须位于出勤表中存在的person_id上（对于之前未显示此内容的道歉！ my BAD 请注意表格右侧的附加列，其中应包含十字形第2周。
2 个答案:
答案 0 :(得分：2)
我个人会经历嵌套数组。
我举个例子。请注意，这只是原则上，因为我没有数据库，我还没有测试过。 :)
1 - ＆gt;数组$liste_name = array();的语句
2 - ＆gt;通过以下方式替换您的第二次治疗：
while( $row = mysql_fetch_assoc($result) ){
$name = $row["given_names"] . " " . $row["surname"];
$weekpk = $row["week_pk"];
$sno = $row["session"];
$person_id = $row["person_id"]; //add person_id
$liste_name[$person_id] = $name; // to have the last name
$tab[$person_id][1] = "<td class='tborder_person_left'>$name</td>"; // to have the last name
for($i=2; $i<=$totalcolumn; $i++){
if( $weeksession[$weekpk][$sno] == $i )
$tab[$person_id][$i] = "<td class='tborder_person_attended'>✔</td>";
}
}//end while
3 - ＆gt;填空区：
foreach ($liste_name as $person_id => $name){
$tab2[$person_id][1] = $tab[$person_id][1];
for($i=2; $i<=$totalcolumn; $i++){
if (!isset($tab[$person_id][$i]) || ($tab[$person_id][$i] != "<td class='tborder_person_attended'>✔</td>"))
$tab[$person_id][$i] = "<td class='tborder_person'>-</td>";
$tab2[$person_id][$i] = $tab[$person_id][$i];
}
}
4 - ＆gt;使用数组：
foreach($tab2 as $person_id => $col){
echo "<tr>";
foreach ($col as $i => $value){
echo $value;
}
echo "</tr>";
}
echo "</table>";
我尝试使用虚构数据并且可以正常运行。 ：）
答案 1 :(得分：2)
我很想使用几个CROSS JOIN来获得会话/周/人的所有可能组合，然后离开加入，以防止出席。
SELECT unique_names.given_names,
unique_names.surname,
week_sessions.week_pk,
week_sessions.session,
attendance.attendance_pk
FROM
(
SELECT week_pk, sub1.i AS session, week_name, week.cohort_fk, cohort.cohort_year
FROM week
INNER JOIN cohort
ON week.cohort_fk = cohort.cohort_pk
INNER JOIN
(
SELECT 1 i UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 10
) sub1
ON week.sessions >= sub1.i
WHERE cohort.cohort_year = 2014
AND cohort.cohort_pk = 1
) week_sessions
CROSS JOIN
(
SELECT DISTINCT given_names, surname
FROM attendance
) unique_names
LEFT OUTER JOIN attendance
ON week_sessions.week_pk = attendance.week_fk
AND week_sessions.cohort_fk = attendance.cohort_fk
AND week_sessions.session = attendance.session
AND unique_names.given_names = attendance.given_names
AND unique_names.surname = attendance.surname
ORDER BY unique_names.given_names,
unique_names.surname,
week_sessions.week_pk,
week_sessions.session
我为此敲了一个SQL小提琴： -
http://www.sqlfiddle.com/#!2/4a388/7
然后你可以轻松地绕过这个（尽管添加标题有点乱 - 以下是没有标题的一半大小）。我已经使用了mysql_ *函数，因为你已经在使用它了。
<?php
$connection = mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("test_area") or die(mysql_error());
$sql = "SELECT unique_names.person_id,
unique_names.FullName,
week_sessions.week_pk,
week_sessions.session,
attendance.attendance_pk
FROM
(
SELECT week_pk, sub1.i AS session, week_name, week.cohort_fk, cohort.cohort_year
FROM week
INNER JOIN cohort
ON week.cohort_fk = cohort.cohort_pk
INNER JOIN
(
SELECT 1 i UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 10
) sub1
ON week.sessions >= sub1.i
WHERE cohort.cohort_year = 2014
AND cohort.cohort_pk = 1
) week_sessions
CROSS JOIN
(
SELECT person_id, MAX(CONCAT_WS(' ', given_names, surname)) AS FullName
FROM attendance
GROUP BY person_id
) unique_names
LEFT OUTER JOIN attendance
ON week_sessions.week_pk = attendance.week_fk
AND week_sessions.cohort_fk = attendance.cohort_fk
AND week_sessions.session = attendance.session
AND unique_names.person_id = attendance.person_id
ORDER BY unique_names.person_id,
unique_names.FullName,
week_sessions.week_pk,
week_sessions.session";
$result = mysql_query($sql, $connection) or die(mysql_error());
$prev_person_id = 0;
$first = true;
$header = array('Name'=>array('Session'));
$output = array();
while( $row = mysql_fetch_assoc($result) )
{
if ($prev_person_id != $row['person_id'])
{
if ($prev_person_id != 0)
{
$first = false;
}
$prev_person_id = $row['person_id'];
$output[$prev_person_id] = array();
}
if ($first)
{
$header["Week ".$row['week_pk']]["S".$row['session']] = "S".$row['session'];
}
$output[$prev_person_id][] = (($row['attendance_pk'] == '') ? ' ' : 'X');
}
$header1 = '';
$header2 = '';
foreach($header as $key=>$value)
{
$header1 .= "<td colspan='".count($value)."'>$key</td>\r\n";
foreach($value as $key1=>$value1)
{
$header2 .= "<td>$value1</td>\r\n";
}
}
echo "<table border='1'>\r\n";
echo "<tr>\r\n$header1</tr>\r\n";
echo "<tr>\r\n$header2</tr>\r\n";
foreach($output as $name=>$value)
{
echo "<tr><td>$name</td>";
foreach($value as $key1=>$value1)
{
echo "<td>$value1</td>\r\n";
}
echo "</tr>";
}
echo "</table>\r\n";
?>
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