如何从一个sql表到另一个sql表添加多行数据，并根据复选框进行选择？

Question

<form method="post">
<?php
    $sql_u="SELECT * from cubaan";
    $query_u = mysqli_query($conn,$sql_u);
    while($row=mysqli_fetch_assoc($query_u)){
?>
        <table border="2">
            <tr>
                <td><input type="checkbox" name="bio[]" value="<?php $row['name'];?>" data-valuetwo="<?php $row['age'];?>" data-valuethree="<?php $row['job'];?>"></td>
                <td><?php echo $row['id'];?></td>
                <td><?php echo $row['name'];?></td>
                <td><?php echo $row['age'];?></td>
                <td><?php echo $row['job'];?></td>
            </tr>
        </table>
<?php
    }
?>
<input type="submit" name="post">
</form>

this is my code where I want to select data from the database and the selected data will insert to the new database

Answer 1

Use input type hidden as following

<form method="post">
<?php
$sql_u="SELECT * from cubaan";
$query_u = mysqli_query($conn,$sql_u);
while($row=mysqli_fetch_assoc($query_u)){
?>
<table border="2">
<tr>
<td>
<input type="checkbox" name="bio[<?php echo $row['id'];?>]" value="<?php echo $row['name'];?>">
<input type="hidden" name="age[<?php echo $row['id'];?>]" value="<?php echo $row['age'];?>">
<input type="hidden" name="job[<?php echo $row['id'];?>]" value="<?php echo $row['job'];?>">
</td>
<td><?php echo $row['id'];?></td>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['age'];?></td>
<td><?php echo $row['job'];?></td>


</tr>
</table>

<?php
}
?>
<input type="submit" name="post">
</form>

And while your form submit action use

<?php foreach($_POST['bio'] as $id=>$name) { 

echo $name;
echo $_POST['age'][$id];
echo $_POST['job'][$id];
}
?>