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Hi everybody and tks in advance for your help! I have a plugin and I was trying to do some pagination for a custom query. Everything seems to work fine, but when I click on page 2 (or whatever not being "1"), wordpress says

You do not have sufficient permissions to access this page

I believe that this error is because with this pagination method that I was trying, the URL is:

/wp-admin/admin.php?page=lismovim-withdraw-slug/page/2/

So, I suppose that WP doesn't reconigze this address. Any ideas? Maybe some other way to do the pagination? I really need some help, I was surfing the internet quite long now and couldn't find any solution. Here's my code:

function listado_movimientos(){


    echo '<p><strong>LISTADO DE MOVIMIENTOS APROBADOS/DECLINADOS<strong></p>';
    echo '<table class="wp-list-table widefat fixed striped axdepositbycash">
    <tr style="font-weight: bold;font-size: 15px">
                <td>ID</td>
                <td>FECHA</td>
                <td>USUARIO</td>
                <td>NOMBRE</td>
                <td>CANTIDAD</td>
                <td>Nº CUENTA</td>
                <td>ESTADO</td>
            </tr>';

    $rows_per_page = 10;
    $current = (intval(get_query_var('paged'))) ? intval(get_query_var('paged')) : 1;

    global $wpdb; 
    $table_name = $wpdb->prefix. "withdraw";
    $result2=$wpdb->get_results("select * from $table_name where status='aprobado'");


    $pagination_args = array(
        'base' => @add_query_arg('paged','%#%'),
        'format' => '',
        'total' => ceil(sizeof($result2)/$rows_per_page),
        'current' => $current,
        'show_all' => false,
        'type' => 'plain',
        );


        $pagination_args['base'] = user_trailingslashit( trailingslashit( remove_query_arg('s',get_pagenum_link(1) ) ) . 'page/%#%/', 'paged');


        if( !empty($wp_query->query_vars['s']) )
        $pagination_args['add_args'] = array('s'=>get_query_var('s'));

        echo paginate_links($pagination_args);

        $start = ($current - 1) * $rows_per_page;
        $end = $start + $rows_per_page;
        $end = (sizeof($result2) < $end) ? sizeof($result2) : $end;

        echo '<br />';
        echo '<br />';

        for ($i=$start;$i < $end ;++$i ) {
        $r2 = $result2[$i];
        echo  '<tr>
                <td>'.$r2->id.'</td>
                <td>'.$r2->wdate.'</td>
                <td> '.$r2->fname. "  " . $r2->lastname.'</td>
                <td>'.$r2->name.'</td>
                <td>'.$r2->amt.'</td>
                <td>'.$r2->acno.'</td>
                <td>'.$r2->status.'</td>
            </tr>';
        }
    echo "</table>";

}

Here is the original snippet, where I found how to do the pagination in the first place.

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The base argument for paginate_links() should be:

admin_url( 'admin.php?page=lismovim-withdraw-slug&paged=%_%' )

... where %_% is replaced by format, which should be:

'%#%'

... which is replaced by the page number.

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  • Tks! It works just fine but now I have another issue. The URL seems to work fine (in the permissions "way") but pagination is not working at all. When I try to see the value of this line: $current = (intval(get_query_var('paged'))) ? intval(get_query_var('paged')) : 1; It's always getting me "1", so... that means that "get_query_var" is not working to read de "paged" parameter... Am I doing something wrong here? Tks again!! – Fryla- Cristian Marucci May 25 '16 at 20:30
  • No, get_query_var works around $wp_query, which is only prepared/ready to use on posts-related admin pages. Use $paged = isset( $_REQUEST['paged']) ? max( 1, ( int ) $_REQUEST['paged'] ) : 1; – TheDeadMedic May 25 '16 at 21:11
  • Oh man, I must say you really really save my day. It works exactly as you said !!! Thanks you very much!!! Hope in the future I can help you with something!! – Fryla- Cristian Marucci May 25 '16 at 21:18

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