获取数组值并根据其自己的列存储它，但值存储在不同的列中

The html checkbox before query, I chose option 1 and option 4

The checkbox value

the data where the values should be stored, but it stores into option 1 and option 2 columns instead while it should store it inside both 'opt1' and 'opt4'

Result inside the table

PHP

if( isset( $_POST['picked'] ) )
        {
            $keepdata = array();

            $keepdata = $_POST['pick'];

            $sqlinsert = "INSERT INTO traincheckbox VALUES ( null, '$keepdata[0]', '$keepdata[1]', '$keepdata[2]', '$keepdata[3]', '$keepdata[4]' )";

            mysqli_query( $sqlnrg, $sqlinsert );

            header( "Location: pick.php" );
        }

HTML

<html>
    <form method="post">
        <input type="checkbox" name="pick[]" value="PAID"><label>Activity 1</label></input>
        <input type="checkbox" name="pick[]" value="PAID"><label>Activity 2</label></input>
        <input type="checkbox" name="pick[]" value="PAID"><label>Activity 3</label></input>
        <input type="checkbox" name="pick[]" value="PAID"><label>Activity 4</label></input>
        <input type="checkbox" name="pick[]" value="PAID"><label>Activity 5</label></input>
        <button type="submit" name="picked">Pick activities</button>
    </form>
</html>

Using name="pick[]" causes only to index checked checkboxes from 0 to n.

If you need explicit indexes, then you should force them:

<input type="checkbox" name="pick[0]" value="PAID"><label>Activity 1</label></input>

and so on...

You should remove $keepdata = array(); line from your code. It does nothing, because then it is initialized by $_POST['pick']. Also have on mind, that values not marked as checked will not be sent in any form. There will be errors because of non-existent keys in array (if you check 1 and 4, then 2 and 3 will not be set). You need to handle this situation.