【发布时间】:2018-04-21 15:56:48
【问题描述】:
我想问一下如何让我的表格显示如下:
但我只能设法让我的表格看起来像这样:
如您所见,表格是根据他们存放东西的不同年份或学期分开的。它基本上是那个特定人的历史存储。不幸的是,我不知道如何将生成的表格附加到同一年和同一学期的表格中,而不是将其分开。甚至编号也会受到影响。以下是我到目前为止的代码:
<?php
include("connect.php");
include("header.php");
if(isset($_GET['invstuview_bag'], $_GET['invstuview_name']))
{
$get_id = $_GET['invstuview_bag'];
$get_name = $_GET['invstuview_name'];
?>
<br/>
<ul class="breadcrumb">
<li class="breadcrumb-item"><a href="view.php">View History</a></li>
<li class="breadcrumb-item-active">Baggage Detail History</a></li>
</ul>
<div id="cssword">Baggage Detail History For <?php echo $get_name; ?>(<?php echo $get_id; ?>)</div>
<div class="container" style="width:70%;">
<div class="table-responsive">
<?php
$sql_join = "SELECT student.*,location.*, inventory.*, baggage.*
FROM
student,location, inventory, baggage
WHERE
student.student_id = inventory.invstu_id AND
baggage.baggage_id = inventory.invbag_id AND
location.location_id = inventory.invloc_id AND
invstu_id = '$get_id'
ORDER BY inventory_id";
$result_join= mysqli_query($connect,$sql_join);
$prev_year = "";
$prev_sem = 0;
$get_year = "";
$get_sem = 0;
while($row_join=mysqli_fetch_assoc($result_join))
{
$counter = 1;
$prev_year = $row_join["invstu_year"];
$prev_sem = $row_join["invstu_sem"];
//if the data is of same year and semester
if(($prev_year!="") && ($prev_sem!=0) && ($prev_year == $get_year) && ($prev_sem == $get_sem))
{
$get_year = $row_join["invstu_year"];
$get_sem = $row_join["invstu_sem"];
?>
<table class="table table-bordered">
<tr>
<th>No.</th>
<th>Baggage Types</th>
<th>Quantity</th>
<th>Location</th>
</tr>
<tr>
<td><?php echo $counter; ?></td>
<td><?php echo $row_join["baggage_type"]; ?></td>
<td><?php echo $row_join["invbag_quantity"]; ?></td>
<td><?php echo $row_join["location_house"]; ?></td>
</tr>
<?php
$counter++;
echo'</table>';
}
//if data is not of same year or semester
else
{
$get_year = $row_join["invstu_year"];
$get_sem = $row_join["invstu_sem"];
?>
</br></br>
Room: <?php echo $row_join["invstu_room"]; ?><br/>
Year: <?php echo $row_join["invstu_year"]; ?><br/>
Semester: <?php echo $row_join["invstu_sem"]; ?><br/>
<table class="table table-bordered">
<tr>
<th>No.</th>
<th>Baggage Types</th>
<th>Quantity</th>
<th>Location</th>
</tr>
<tr>
<td><?php echo $counter; ?></td>
<td><?php echo $row_join["baggage_type"]; ?></td>
<td><?php echo $row_join["invbag_quantity"]; ?></td>
<td><?php echo $row_join["location_house"]; ?></td>
</tr>
<?php
$counter++;
echo'</table>';
}
}
?>
</div>
<div align="right">
<button type="button" name="back" class="btn btn-success" id="back" style="width:200px; display:inline-block; margin-top:2%; margin-bottom:1%; background-color: #4CAF50" onclick="window.location.href='view.php'">Back</button>
</div>
</div>
<?php
}
?>
非常感谢任何想法。
【问题讨论】:
-
嗨。请use text, not images/links, for text (including code, tables & ERDs)。使用图像仅是为了方便补充文本和/或文本中无法提供的内容。永远不要给出没有图例/键的图表。请阅读minimal reproducible example并采取行动。
标签: php mysql html-table