pivot 相关问题

我有一个包含数据的学术表。例如：等级百分比程度 50 文凭 70 文凭后 67 这是我的代码，但不起作用：学位&l... 我有一个包含数据的学术表。例如：级别百分比学位 50 文凭 70 文凭后 67 这是我的代码，但不起作用： <table> <tr> <th>Degree</th> <th>Diploma</th> $stmtc = $conn->prepare("SELECT MAX(CASE WHEN level = 'Degree' THEN percentage END) AS $degree, MAX(CASE WHEN level = 'Diploma' THEN percentage END) AS $diploma FROM tbl_academic_qualification WHERE member_no = '$emp_id'"); $stmtc->execute(); $resultc = $stmtc->fetchAll(); foreach($resultc as $rowc){ $degree = $rowc['percentage']; $diploma = $rowc['percentage']; ?> <tr> <td><?php echo "$degree"; ?></td> <td><?php echo "$diploma"; ?></td> </tr> </table> 我想这样显示：学位文凭文凭后 50 70 67 试试这个： <table border="1"> <tr> <th>Degree</th> <th>Diploma</th> <th>Post Diploma</th> </tr> <?php $stmt = $conn->prepare("SELECT MAX(CASE WHEN level = 'Degree' THEN percentage END) AS Degree, MAX(CASE WHEN level = 'Diploma' THEN percentage END) AS Diploma, MAX(CASE WHEN level = 'Post Diploma' THEN percentage END) AS Post_Diploma FROM tbl_academic_qualification WHERE member_no = :emp_id"); $stmt->execute(); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { echo "<tr>"; echo "<td>" . $row['Degree'] . "</td>"; echo "<td>" . $row['Diploma'] . "</td>"; echo "<td>" . $row['Post_Diploma'] . "</td>"; echo "</tr>"; } ?> </table>

php html pivot fetch

回答 1 投票 0

需要按一天中的每小时进行更广泛的调整，而无需创建多个日条目

我有一个数据集，其中包含一年中每一天的每小时空气质量数据。我正在尝试扩大范围，但当 NULLCODE 列中存在空代码时遇到问题。这些特殊的日子

r pivot tidyr

回答 1 投票 0

我如何在php中将行数据转换为列，请帮助我

我有一个包含数据的学术表。请帮我解决这个问题例如：等级百分比程度 50 文凭 70 文凭后 67 这是我的代码，但不起作用 ... 我有一个包含数据的学术表。请帮我解决这个问题例如：级别百分比学位 50 文凭 70 文凭后 67 这是我的代码，但不起作用 <table> <tr> <th>Degree</th> <th>Diploma</th> $stmtc = $conn->prepare("SELECT MAX(CASE WHEN level = 'Degree' THEN percentage END) AS $degree, MAX(CASE WHEN level = 'Diploma' THEN percentage END) AS $diploma FROM tbl_academic_qualification WHERE member_no = '$emp_id'"); $stmtc->execute(); $resultc = $stmtc->fetchAll(); foreach($resultc as $rowc){ $degree = $rowc['percentage']; $diploma = $rowc['percentage']; ?> <tr> <td><?php echo "$degree"; ?></td> <td><?php echo "$diploma"; ?></td> </tr> </table> 我想这样展示学位文凭文凭后 50 70 67 试试这个： <table border="1"> <tr> <th>Degree</th> <th>Diploma</th> <th>Post Diploma</th> </tr> <?php $stmt = $conn->prepare("SELECT MAX(CASE WHEN level = 'Degree' THEN percentage END) AS Degree, MAX(CASE WHEN level = 'Diploma' THEN percentage END) AS Diploma, MAX(CASE WHEN level = 'Post Diploma' THEN percentage END) AS Post_Diploma FROM tbl_academic_qualification WHERE member_no = :emp_id"); $stmt->execute(); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { echo "<tr>"; echo "<td>" . $row['Degree'] . "</td>"; echo "<td>" . $row['Diploma'] . "</td>"; echo "<td>" . $row['Post_Diploma'] . "</td>"; echo "</tr>"; } ?> </table>

php html pivot fetch

回答 1 投票 0

pivot 相关问题

最新问题