将选中的项插入另一个sql表？（PHP和SQL）

Hi everyone I'm new here ! I'm wondering how I can insert the values of a printed row when it's specific checkbox is ticked.

here is my code:

<?php
$query = "SELECT * FROM phpmyreservation_forvalidation";
$result = mysql_query($query);
?>

<form action='insertrecords.php' method='post'>
<?php
echo "<table  cellpadding='0' cellspacing='0' border='1'>";
echo "<tr> <th>reservation_id</th> <th>reservation_user_name</th> <th>reservation_day</th> <th>reservation_week</th> 
    <th> # </th></tr>";
while($row = mysql_fetch_array( $result )) {
echo "<tr><td>"; 
echo $row['reservation_id'];   
echo "</td><td>"; 

echo $row['reservation_user_name'];
echo "</td><td>"; 

echo $row['reservation_day'];
echo "</td><td>"; 

echo $row['reservation_week'];
echo "</td><td align='center'>"; 

echo "</td>";  
?>

<td><input name="checkbox[]" type="checkbox" id="checkbox[]"  value="<?= $row['id'] ?>" ></td>

<?php
} 
echo "</table>";
?>


<input type='submit' value='Submit' />
</form>

I'm not sure if what i've done is correct, I also don't have an insert records.php yet, I tried to make one but it didn't work and now I'm really confused. Does anyone know how I can do this ? Thank you !

additional notes: What I'm doing is a validation system the schedules that are not yet validated are on a separate table, when the admin validates them it gets inserted into the phpmyreservation_reservations table.

Thanks !

fetch the 'phpmyreservation_forvalidation' result into variables and insert it into your 'phpmyreservation_reservations' table when the form is submit.

In records.php:

$row_id=$_POST['checkbox'];
foreach ($row_id as $key=>$val)
{
    //do insert here with $val
}

If I were you, I would use AJAX calls to obtain all the data needed from your database and also to insert the data to a new table. In the example below I used jQuery to do ajax calls.

my example web page code:

<html>
<head>
    <meta charset="utf-8" />
    <meta name="viewport" content="width=device-width" />
    <title>Title of the page</title>
    <script src="jquery-1.10.2.min.js"></script>

    <script>
    // Function beeing called when a checkbox is changed
    function MyCheckboxChanged(sender)
    {

      // Send the data to be processed to the server if the checkbox is checked
      if (sender.checked)
      {
        var MyObject            = {};
        MyObject.ValueToInsert  = sender.value;

        DoAJAXData("insert", MyObject);
      }
    }

    // Function responsible for AJAX calls - requires jQuery
    function DoAJAXData(command, MyObject)
    {
      $.post("phpReceiver.php",
      {
        command:command,
        sendObject:JSON.stringify(MyObject)
      })
      .success(function(data)
      {
        // Process the data echoed from php
        // In my case, should  alert: "successfully inserted data: " + your selected checkbox value
        alert(data);

      })
      .fail(function(error){
        //alert("Unable to retrieve data from the server");
      });
    }

    </script>

</head>
<body>
<?php

for ($i=0; $i < 5; $i++) 
{ 
  $var = <<<CHB
  <input type="checkbox" onchange="MyCheckboxChanged(this);" name="$i" value="Bike number $i">I have a bike number $i</br>
CHB;
  echo $var;
}
?>

</body> 
</html>

and the phpReceiver.php code:

<?php

$command    = $_POST['command'];
$object     = json_decode($_POST['sendObject'], true);

if ($command == "insert") 
{
    $valueToInsert = $object['ValueToInsert'];

    // Do you insertion to the database here

    // Echo any results if wanted
    echo "successfully inserted data: $valueToInsert";

}
?>

将选中的项插入另一个sql表？ （PHP和SQL）

将选中的项插入另一个sql表？（PHP和SQL）