<?php
$con = mysqli_connect('localhost', 'root', '');
if(!$con)
{
die("not ok");
}
mysqli_select_db($con,"uoh");
$q1 = "SELECT * FROM student_record INNER JOIN degree_plan ON
student_record.course_number = degree_plan.course_number
INNER JOIN courses ON student_record.course_number =
courses.course_number where student_record.id = 201102887 AND degree_plan.major='COE'";
$result = mysqli_query($con , $q1 ) ;
$data = array();
while($row = mysqli_fetch_array($result))
{
$data[$row["term_no"]][] = array(
'code' => $row["code"],
'grade' => $row["grade"]
);
}
echo '<table width="200" border="1">';
echo "<tr>";
echo "<th>courses</th>";
echo "<th>terms</th>";
echo "<th>grades</th>";
echo "</tr>";
foreach($data as $term=>$otherrow) {
$count = 0;
foreach ($otherrow as $data) {
if($count == 0) {
echo "<tr>";
echo "<td>" . $data["code"]. "</td>";
echo '<td rowspan="'.count($otherrow).'">' . $term. '</td>';
echo "<td>" . $data["grade"]. "</td>";
echo "</tr>";
}
else
{
echo "<tr>";
echo "<td>" . $data["code"]. "</td>";
echo "<td>" . $data["grade"]. "</td>";
echo "</tr>";
}
$count++;
}
}
echo "</table>";
?>
我有这段代码，它工作得很好，但是当我想添加更多列时遇到了问题。我尝试添加第四列(echo "<td>" . $row["crd"]. "</td>";(，但没有结果.它给了我空的单元格。我怎么能做到这一点？
我想将此回显"<td>" . $row["crd"]. "</td>";列添加到我的代码中。
如评论中所述，已注意到两个错误。
- 您将在第二个 foreach 循环中重新声明
$data - 您没有在任何地方启动
$row，并且试图回显$row["crd"]将导致一个空单元格。
建议的解决方案：
将 foreach 循环中 $data 值的名称更改为 $row，从而同时解决这两个问题：
foreach($data as $term=>$otherrow) {
$count = 0;
foreach ($otherrow as $row) {
if($count == 0) {
echo "<tr>";
echo "<td>" . $row["code"]. "</td>";
echo '<td rowspan="'.count($otherrow).'">' . $term. '</td>';
echo "<td>" . $row["grade"]. "</td>";
echo "</tr>";
}
else
{
echo "<tr>";
echo "<td>" . $row["code"]. "</td>";
echo "<td>" . $row["grade"]. "</td>";
echo "</tr>";
}
$count++;
}
}
当您现在添加echo "<td>" . $row["crd"]. "</td>";时，它应该回显存储在$row数组中的值(当然，只要该值首先是从数据库中的表中提取的(。
让我知道这是否对您有用。