还有其他方法可以像矩阵形式一样在 html 中打印 mysql table 吗?
Is there any other way to print mysql table in html like a matrix form?
我目前在一所大学工作project.The要求是将数据(主题,class,部分)打印到html [=32=中相应的日期和时间] 尽力达到要求,但 html table 只是再次打印并打印下一个值,我希望我的所有数据在单个 table 中打印到相应的日期和时间.
这是我的数据库的图像:
这是我不满意的输出:
这是我的查询:
<?php
include "conn.php";
include ("/styles/faculty_menu.php");
session_start();
$uid=$_SESSION['userid'];
$cuid=strtoupper("$uid");
$qrun=mysql_query("SELECT * FROM login WHERE id='$cuid'");
$udata=mysql_fetch_assoc($qrun);
$q2run=mysql_query("SELECT * FROM schedule WHERE id='$cuid'");
$u2data=mysql_fetch_assoc($q2run);
$urole=$udata['role'];
$comp=strtoupper("$u2data[subject]");
$day = array("Monday", "Tuesday", "Wednessday", "Thursday", "Friday", "Saturday");
?>
我希望输出是单个 table,所有受尊重的数据都在相应的位置。
这是我的代码:
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<th>
<center>Day/Hour</center>
</th>
<?php
for ($i=1; $i <=8 ; $i++) {
?>
<th>
<center><?php echo $i; ?></center>
</th>
<?php } ?>
</tr>
<?php
while ($rows=mysql_fetch_assoc($q2run)) {
for ($d=0; $d <6 ; $d++) {
?>
<tr>
<th>
<center><?php echo $day[$d]; ?></center>
</th>
<?php
for ($r=1; $r <=8; $r++) {
?>
<td>
<?php
if ($rows['day']==$day[$d] && $rows['hour']==$r) {
echo $rows['class'],$rows['section'],$rows['subject'];
}
?>
</td>
<?php } ?>
</tr>
<?php }
}
?>
</table>
我已经更改了您的代码的一些逻辑。希望这会奏效
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<th>
<center>Day/Hour</center>
</th>
<?php
for ($i=1; $i <=8 ; $i++)
echo '<th><center>'.$i.'</center></th>';
?>
</tr>
<?php
for($d=0;$d<6;$d++){
?>
<tr>
<th>
<center><?php echo $day[$d]; ?></center>
</th>
<?php
for ($r=1; $r <=8; $r++) {
echo '<td>';
while ($rows=mysql_fetch_assoc($q2run)) {
if($rows['day']==$day[$d] && $rows['hour']==$r)
echo $rows['class'],$rows['section'],$rows['subject'];
}
echo '</td>';
}
?>
</tr>
<?php
}
?>
</table>
</div>
我目前在一所大学工作project.The要求是将数据(主题,class,部分)打印到html [=32=中相应的日期和时间] 尽力达到要求,但 html table 只是再次打印并打印下一个值,我希望我的所有数据在单个 table 中打印到相应的日期和时间.
这是我的数据库的图像:
这是我不满意的输出:
这是我的查询:
<?php
include "conn.php";
include ("/styles/faculty_menu.php");
session_start();
$uid=$_SESSION['userid'];
$cuid=strtoupper("$uid");
$qrun=mysql_query("SELECT * FROM login WHERE id='$cuid'");
$udata=mysql_fetch_assoc($qrun);
$q2run=mysql_query("SELECT * FROM schedule WHERE id='$cuid'");
$u2data=mysql_fetch_assoc($q2run);
$urole=$udata['role'];
$comp=strtoupper("$u2data[subject]");
$day = array("Monday", "Tuesday", "Wednessday", "Thursday", "Friday", "Saturday");
?>
我希望输出是单个 table,所有受尊重的数据都在相应的位置。
这是我的代码:
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<th>
<center>Day/Hour</center>
</th>
<?php
for ($i=1; $i <=8 ; $i++) {
?>
<th>
<center><?php echo $i; ?></center>
</th>
<?php } ?>
</tr>
<?php
while ($rows=mysql_fetch_assoc($q2run)) {
for ($d=0; $d <6 ; $d++) {
?>
<tr>
<th>
<center><?php echo $day[$d]; ?></center>
</th>
<?php
for ($r=1; $r <=8; $r++) {
?>
<td>
<?php
if ($rows['day']==$day[$d] && $rows['hour']==$r) {
echo $rows['class'],$rows['section'],$rows['subject'];
}
?>
</td>
<?php } ?>
</tr>
<?php }
}
?>
</table>
我已经更改了您的代码的一些逻辑。希望这会奏效
<div class="table-responsive">
<table class="table table-bordered">
<tr>
<th>
<center>Day/Hour</center>
</th>
<?php
for ($i=1; $i <=8 ; $i++)
echo '<th><center>'.$i.'</center></th>';
?>
</tr>
<?php
for($d=0;$d<6;$d++){
?>
<tr>
<th>
<center><?php echo $day[$d]; ?></center>
</th>
<?php
for ($r=1; $r <=8; $r++) {
echo '<td>';
while ($rows=mysql_fetch_assoc($q2run)) {
if($rows['day']==$day[$d] && $rows['hour']==$r)
echo $rows['class'],$rows['section'],$rows['subject'];
}
echo '</td>';
}
?>
</tr>
<?php
}
?>
</table>
</div>