【发布时间】:2015-10-01 03:58:37
【问题描述】:
PHP 代码:
<?php
//DB CONNECTION
//---(1)Get skillname---
$q = "SELECT skillName FROM skill ORDER BY skillName asc";
$r = mysqli_query($dbc, $q);
$num_rows = mysqli_num_rows($r);
while($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
$skills[] = $row['skillName'];
}
//---(2)Get classname---
$q1 = "SELECT className FROM class";
$r1 = mysqli_query($dbc, $q1);
$num_rows1 = mysqli_num_rows($r1);
while($row1 = mysqli_fetch_array($r1, MYSQLI_ASSOC))
{
$className[] = $row1['className'];
}
//---(3)Create table---
echo '<table border="1" style="border-collapse: collapse; text-align: center">';
echo '<tr>';
for($a = 0; $a < $num_rows; $a++)
{
echo '<th colspan="2">'.$skillName[$a].'</th>';
}
echo '</tr>';
for($b = 0; $b < $num_rows; $b++)
{
echo '<th>Student Name</th>';
echo '<th>Grade</th>';
}
echo '</tr>';
//---(4)Get student name and grade---
for($s = 0; $c < $num_rows1; $c++)
{
$q2 = "SELECT GROUP_CONCAT(sm.studentName) as studentName,
GROUP_CONCAT(sm.studentGrade) as studentGrade,
s.skillName
FROM studentmark sm
LEFT JOIN skill s ON sm.skillID = s.skillID
WHERE sm.className = '".$className[$c]."'
GROUP BY s.skillID";
$r2 = mysqli_query($dbc, $q2);
$num_rows2 = mysqli_num_rows($r2);
$value = array();
while($row2 = mysqli_fetch_array($r2, MYSQLI_ASSOC))
{
$value[] = $row2;
}
echo '<tr>';
for($d = 0; $d < $num_rows2; $d++)
{
echo '<td>'.$value[$d]['studentName'].'</td>';
echo '<td>'.$value[$d]['studentGrade'].'</td>';
}
echo '</tr>';
}
echo '</table>';
?>
从上面的代码,我的输出如下:
我快完成了。我可以在 1 行中显示学生姓名和年级。
现在,我最不想做的就是将它们放入合适的技能名称中,如下所示:
我想比较$q2 上的$skills 和s.skillname。
以下是我的逻辑:
if($value[X]['skillName'] == $skills[X])
{
//put student name and grade inside
}
else
{
//empty cell
}
但我不知道我应该在哪里打开 for 循环并将我的逻辑放在 (4) 中。有人可以帮我吗?
【问题讨论】: