我为项目留出时间和时间，但我陷入了这个
如何将其显示为php？
名称|中间名|姓氏| [DTR]（DTR是指向dtr.php的其他页面的链接）
然后进入[DTR]链接，它将显示
时间在| TIME_OUT
[添加时间]（[添加时间]是add_time.php的链接所以基本上时间和时间是dtr.php中的循环
问题是我可以添加time_in和time_out也在我提交后我不能直接转到dtr.php？= $ id
到目前为止，这是我尝试过的一个
view.php
<?php
include 'dbconfig.php';
$sql = "SELECT * FROM info";
$result = mysql_query($sql);
?>
"enter code here"
while ($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td><center>".$row['firstname']."</center></td>";
echo "<td><center>".$row['middlename']."</center></td>";
echo "<td><center>".$row['lastname']."</center></td>";
echo "<td><a href= \"dtr.php?id=".$row['id']." \"> DTR";
echo "</tr>";
}
dtr.php
<?php
include 'dbconfig.php';
$time_id = $_GET['id'];
$sql = "SELECT * FROM time WHERE id = '$time_id'";
$result = mysql_query($sql);
?>
"enter code here"
<?php
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td><center>".$row['time_in']."</center></td>";
echo "<td><center>".$row['time_out']."</center></td>";
echo "<td><a href= \"add_time.php?id=".$row['id']." \"> edit </a></td>";
echo "</tr>";
}
?>
add_time.php
<?php
include 'dbconfig.php';
$time_id = $_GET['id'];
$sql = "SELECT * FROM time WHERE id = '$time_id'";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
?>
<form action="insert_time.php" method="POST">
time in: <input type="text" name="time_in"/><br/>
time out: <input type="text" name="time_out"/><br/>
</form>
<?php echo '<td><form action="dtr.php?id=' . $row['id'] . '"><input type="submit" name="save" value="Save" /></form></td>'; ?>
insert_time.php
<?php
include 'dbconfig.php';
$time_id = $_GET['id'];
$sql = "SELECT * FROM time WHERE id = '$time_id'";
$result = mysql_query($sql);
$row = mysql_fetch_array($result);
$time_in = $_POST['time_in'];
$time_out = $_POST['time_out'];
$sql = "INSERT INTO time (time_in, time_out) VALUES ('$time_in','$time_out')
$query = mysql_query($sql);
mysql_close();
Header("Location: dtr.php?id=".$time_id);
?>
这是我的信息表
| id | firstname | middlename | lastname |
时间表
| time_id | id | time_in | time_out |