问题描述
我创建了时间和超时系统.这是表
I created a time in and time out system. Here is the table
时间表
| id | time id | time in | time out |
| 1 | 1 | 1:00 am | |
| 1 | 2 | 1:013am | |
id 是外键,而 time id 是时间表 主键.如何更新超时表?
id is a foreign key while the time id is a time table primary key. How can i update the timeout table?
这是我时间
$id = $_GET['id'];
$timein = date('Y/m/d H.i.s');
$sql = "INSERT INTO time (id, timein)
VALUES
('$id','$timein')
";
$query = mysql_query($sql);
mysql_close();
if(is_resource($query) and mysql_num_rows($query)>0){
$row = mysql_fetch_array($query);
echo $row["id"];
}
echo $sql;
这是正确的,但我的问题是如何更新超时?
which is correct, but my question is how can I update the time out?
这将针对特定的时间 ID,例如时间 ID 1.但我也想更新超时.
This will target the certain time id, for example time id 1. But I also want to update time out.
我已经尝试过这样的事情
I already tried something like this
$timeid = $_SESSION['timeid'];
$id = $_GET['id'];
$timein = $_POST['timein'];
$timeout = date('Y/m/d H.i.s');
$sql = "UPDATE time SET(time_id, id, timein, timeout)
VALUES
('$time_id','$id','$timein','$timeout') WHERE id = $id
";
$query = mysql_query($sql);
mysql_close();
if(is_resource($query) and mysql_num_rows($query)>0){
$row = mysql_fetch_array($query);
echo $row["id"];
}
echo $sql;
这是用于查看时间和超时上下文的代码,那么我如何获得确切的时间 ID 来编辑它?
and this is the code is use to view the context of time in and time out so how can i get the exact timeid to edit it?
include_once 'dbconfig.php';
session_start();
$id = $_GET['id'];
$sql = "SELECT * FROM time WHERE id = $id";
$result = mysql_query($sql);
?>
while($row = mysql_fetch_array($result)){
echo "<tr>";
echo "<td><center>".$row['time_id']."</center></td>";
echo "<td><center>".$row['timein']."</center></td>";
echo "<td><center>".$row['timeout']."</center></td>";
echo "<td><center>".$row['totalmin']."</center></td>";
echo "<td><a href= \"timeout.php?id=".$row['time_id']." \"> timeout";
echo "</tr>";
多亏了 MAWIA,更新没问题,但是当我更新时,只有一个 timeid 会更新
the update is ok thanks to MAWIA but when i update only one timeid will get updated
推荐答案
如果你想更新超时,我认为你不需要更新 time_id、id 和 timein,因为你已经在你的第一个语句中插入了.所以你的更新语句是:
If you want to update a timeout, I think you don't need to update time_id, id and timein because you already insert in your first statement. So your update statement would be:
$checkFk = mysql_query("SET FOREIGN_KEY_CHECKS=0");//turn off foreign key
$sql = "UPDATE `time` SET `timeout`='$timeout' WHERE `timeid` = '$timeid'";
$resetFk = mysql_query("SET FOREIGN_KEY_CHECKS=1");//turn on foreign key
Don't use mysql extension Please look at mysql improved extension or pdo
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