我正在尝试使用Wordpress paginate功能最多只显示3页（客户端希望这样），这样当您进入页面时，它会显示您当前所在的页面以及上一页和下一页：

不幸的是，当我这样做时，我得到了几页，包括最后一页：

这是我当前的分页代码，请注意格式是这样的，因为我有选项卡，每个选项卡都需要用不同的查询分页

echo \'<div id ="nav_pages">\'; 
    echo \'<div class="prev_first"></div>\'; 
    echo \'<div class="pages">\';
$pag_args1 = array(
\'type\'         => \'list\',
\'prev_next\'       => False,
\'end_size\'        => 1,
\'add_args\'        => false,
\'add_fragment\'    => \'\',
\'show_all\' => false,
\'base\'         => \'\' . $url . \'?paged1=%#%\',                
    \'format\'  => \'?paged1=%#%\',
    \'current\' => $paged1,
    \'total\'   => $get_mining->max_num_pages);

    echo paginate_links( $pag_args1 );
    echo \'</div>\';
    echo \'<div class="page_x_of_y2">Page <span>\' . $paged1 . \'</span> of <span>\' . $get_mining->max_num_pages . \'</span></div>\';    
echo \'</div>\';

我是否可以修改此代码，使其看起来像第一个图像中的分页，而不是第二个图像中的分页？我是否需要编写一个自定义分页函数来实现这一点？我不同意客户的要求，因为我觉得后者是更好的选择，但你知道他们说什么：“客户总是对的”

UPDATED PARTIAL SOLUTION

在做了更多的研究之后，我发现了以下链接：Getting paginate_links()\'end_size\' to display 0这个解决方案几乎可以让我实现我想要的目标，下面是我修改后的代码：

echo \'<div id ="nav_pages">\'; 
    echo \'<div class="prev_first"></div>\'; 
    echo \'<div class="pages">\';
$pag_args1 = array(
\'type\'         => \'array\',
\'prev_next\'       => False,
\'end_size\'        => 0,
\'mid_size\'        => 1,
\'add_args\'        => false,
\'add_fragment\'    => \'\',
\'show_all\' => false,
\'base\'         => \'\' . $url . \'?paged1=%#%\',                
    \'format\'  => \'?paged1=%#%\',
    \'current\' => $paged1,
    \'total\'   => $get_mining->max_num_pages);
    $paginate_links = paginate_links( $pag_args1 );
    $c=$pag_args1[\'current\'];
    $allowed=[sprintf(\'/?paged1=%d\',$c-1),\'current\',sprintf(\'?paged1=%d\',$c+1)];
    $paginate_links=array_filter(
        $paginate_links,
    function( $value ) use ( $allowed ) {
        foreach( (array) $allowed as $tag )
        {
            if( false !== strpos( $value, $tag ) )
                return true;
        }
        return false;
    }
);

   if( ! empty( $paginate_links ) )
        printf("<ul class=\'page-numbers\'>\\n\\t<li>%s</li>\\n</ul>\\n",join( "</li>\\n\\t<li>", $paginate_links));
    echo \'</div>\';
    echo \'<div class="page_x_of_y2">Page <span>\' . $paged1 . \'</span> of <span>\' . $get_mining->max_num_pages . \'</span></div>\';    
    echo \'</div>\';


}//Endif

这几乎达到了我想要的，现在唯一的问题是，当我超过第1页时，它不会显示上一页。我越来越近了！！

echo \'<div id ="nav_pages">\'; 
    echo \'<div class="prev_first"></div>\'; 
    echo \'<div class="pages">\';
$pag_args1 = array(
\'type\'         => \'array\',
\'prev_next\'       => False,
\'end_size\'        => 0,
\'mid_size\'        => 1,
\'add_args\'        => false,
\'add_fragment\'    => \'\',
\'show_all\' => false,
\'base\'         => \'\' . $url . \'?paged1=%#%\',                
    \'format\'  => \'?paged1=%#%\',
    \'current\' => $paged1,
    \'total\'   => $get_mining->max_num_pages);
    $paginate_links = paginate_links( $pag_args1 );
    $c=$pag_args1[\'current\'];
    $allowed=[sprintf(\'?paged1=%d\',$c-1),\'current\',sprintf(\'?paged1=%d\',$c+1)];
    $paginate_links=array_filter(
        $paginate_links,
    function( $value ) use ( $allowed ) {
        foreach( (array) $allowed as $tag )
        {
            if( false !== strpos( $value, $tag ) )
                return true;
        }
        return false;
    }
);

   if( ! empty( $paginate_links ) )
        printf("<ul class=\'page-numbers\'>\\n\\t<li>%s</li>\\n</ul>\\n",join( "</li>\\n\\t<li>", $paginate_links));
    echo \'</div>\';
    echo \'<div class="page_x_of_y2">Page <span>\' . $paged1 . \'</span> of <span>\' . $get_mining->max_num_pages . \'</span></div>\';    
    echo \'</div>\';

.dots ~ .dots + .page-numbers { display: none; }