Question

这里提供的代码不是用于制作而是开发，我有一个php页面如下：

mysql_select_db($database_dd, $dbs);
$query_Getuser = sprintf("SELECT * FROM users, firstname, lastname WHERE status = 'active'",
$Getuser = mysql_query($query_Getuser, $dd) or die(mysql_error());
$row_Getuser = mysql_fetch_assoc($Getuser);
$totalRows_Getuser = mysql_num_rows($Getuser);

if($Getuser){
    // insert my my output values into my new table
}

<table class="table table-responsive">
  <tr>
    <td>User</td>
    <td><?php echo $row_Getuser['username']; ?></td>
  </tr>
  <tr>
    <td>Status</td>
    <td><?php echo $row_Getuser['status']; ?></td>
  </tr>
  <tr>
    <td>Total Active Users</td>
    <td><?php echo $totalRows_Getuser ?></td>
  </tr>
  <tr>
    <td>Total Inactive Users </td>
    <td><?php echo $totalRows_WScore ?></td>
  </tr>
  </table>

问题：我厌倦了捕获输出并将其重新插入到我在数据库中创建的新表中。我尝试了隐藏字段，但这需要一个页面按钮isset()然后我教会了如何使用Session_start()数据：

$_SESSION['username'] = $_POST['username'];
$_SESSION['status'] = $_POST['status'];
$_SESSION['totalRows_Getuser'] = $_POST['totalRows_Getuser'];

我似乎无法插入数据。可以做些什么？

Answer 1

你可以在

时完成

while($row = mysql_fetch_assoc($Getuser)){

   .... <?= $row_Getuser['username'] ?>
        <?= $row_Getuser['status'] ?>

}

我更希望让MySQL计数。

SELECT COUNT(users) AS ActiveUsers,* FROM users, firstname, lastname WHERE status = 'active'

Answer 2

if both table have same table schema you can use below query

INSERT INTO newtable
SELECT * FROM users

OR if you want to just copy selected columns data into another table you can use.

INSERT INTO Customers (CustomerName, Country)
SELECT SupplierName, Country FROM Suppliers

将php输出重新插入数据库

2 个答案: