Question

我制作了这段代码，并希望这样做，如果我选择另一个选择器中的任何一个不要重置为起始值。我似乎无法弄清楚我的代码出错了。有什么想法吗？

    <?php
      $filter = true;
      $order = "&orderby=post_date&order=DESC";
      if ($sort1 == '' && $sort2 == '' && $sort3 == '') { $sort1 = ' selected="selected"'; }
      if ($_POST['select1'] == 'newest') { $order = "&orderby=post_date&order=DESC"; $sort1 = ' selected="selected"'; $sort2 = ''; $sort3 = ''; }
      if ($_POST['select1'] == 'oldest') { $order = "&orderby=post_date&order=ASC"; $sort2 = ' selected="selected"'; $sort1 = ''; $sort3 = ''; }
      if ($_POST['select1'] == 'most-popular') { $order = "&meta_key=post_views_count&orderby=meta_value_num&order=DESC"; $sort3 = ' selected="selected"'; $sort1 = ''; $sort2 = ''; }

      if ($view1 == '' && $view2 == '' && $view3 == '') { $view1 = ' selected="selected"'; }
      if ($_POST['select2'] == 'list') { $view1 = ' selected="selected"'; $view2 = ''; }
      if ($_POST['select2'] == 'thumbnail') { $view2 = ' selected="selected"'; $view1 = ''; }
    ?>
    <?php if ($filter) { ?>
    <div class="secondNavCategory">
        <nav class="categorymenu">
    <div class="leftNavCategory">
    <form method="post" id="order">
      SORT:
      <select name="select1" onchange='this.form.submit()'>
        <option value="newest"<?=$sort1?>>NEWEST</option>
        <option value="oldest"<?=$sort2?>>OLDEST</option>
        <option value="most-popular"<?=$sort3?>>MOST POPULAR</option>
      </select>
    </form>
    </div>

    <div class="leftNavCategory">
    <form method="post" id="order">
      VIEW:
      <select name="select2" onchange='this.form.submit()'>
        <option value="list"<?=$view1?>>LIST</option>
        <option value="thumbnail"<?=$view2?>>THUMBNAIL</option>
      </select>
    </form>
    </div>
        </nav>
    </div> <!-- secondNavCategory -->
<?php $filter = false; ?>
<?php } ?>

Answer 1

首先你要改变第二种形式的id。大多数浏览器都会获得第一个可用的DOM元素（在您的情况下，第一个表单是因为您的表单ID都相同）。由于您基于onChange事件提交表单，它将正常工作，但请尽量遵循标准。

您必须使用Ajax表单提交并相应地处理结果，因此不会刷新整个页面，并且您不会丢失第二个表单选定值。

此致 HBKsagar

Answer 2

在您的代码中，您忘记关闭if（）{}

</div> <!-- secondNavCategory -->
<?php } ?> <!-- You are missing this line -->

你也在使用未定义的变量，检查它们是否用isset（$ var）设置，如果没有则给它们一个默认值。

除了我得到的未定义变量的通知，它对我有用。

编辑：这是有效的，问题是你有2个表单，如果提交了一个表单，其他的值都没有提交，每个提交的表单只传递它的值，我现在只制作一个表单，如果你想要两个表单，使用彼此的值进行隐藏输入，并使用JS进行设置。或者使用AJAX

<?php
  $filter = true;
  $sort1 = '';
  $sort2 = '';
  $sort3 = '';

  $view1 = '';
  $view2 = '';
  $view3 = '';

  if (isset($_POST['select1'])) $select1 = $_POST['select1'];
  else $select1 = '';

  if (isset($_POST['select2'])) $select2 = $_POST['select2'];
  else $select2 = '';

  $order = "&orderby=post_date&order=DESC";
  if ($sort1 == '' && $sort2 == '' && $sort3 == '') { $sort1 = ' selected="selected"'; }
  if ($select1 == 'newest') { $order = "&orderby=post_date&order=DESC"; $sort1 = ' selected="selected"'; $sort2 = ''; $sort3 = ''; }
  if ($select1 == 'oldest') { $order = "&orderby=post_date&order=ASC"; $sort2 = ' selected="selected"'; $sort1 = ''; $sort3 = ''; }
  if ($select1 == 'most-popular') { $order = "&meta_key=post_views_count&orderby=meta_value_num&order=DESC"; $sort3 = ' selected="selected"'; $sort1 = ''; $sort2 = ''; }

  if ($view1 == '' && $view2 == '' && $view3 == '') { $view1 = ' selected="selected"'; }
  if ($select2 == 'list') { $view1 = ' selected="selected"'; $view2 = ''; }
  if ($select2 == 'thumbnail') { $view2 = ' selected="selected"'; $view1 = ''; }
?>
<html>
    <head><title>My title</title></head>
    <body>
        <?php if ($filter) { ?>
        <div class="secondNavCategory">
            <nav class="categorymenu">
                <form method="post" id="order">
                    <div class="leftNavCategory">
                        SORT:
                        <select name="select1" onchange='this.form.submit()'>
                            <option value="newest"<?php echo $sort1; ?> >NEWEST</option>
                            <option value="oldest"<?php echo $sort2; ?> >OLDEST</option>
                            <option value="most-popular"<?php echo $sort3; ?> >MOST POPULAR</option>
                          </select>
                        </div>
                    <div class="leftNavCategory">
                        VIEW:
                        <select name="select2" onchange='this.form.submit()'>
                             <option value="list"<?php echo $view1; ?> >LIST</option>
                            <option value="thumbnail"<?php echo $view2; ?> >THUMBNAIL</option>
                         </select>
                    </div>  
                </form>
            </nav>
        </div> <!-- secondNavCategory -->
        <?php } ?>
    </body>
</html>

如何使两个选择器保持其值

2 个答案: