Question

我的状态表有state_id和state_name。我将它直接显示在模板文件中，如下所示..

  <select name="state"  id="state"  class="select-submit2">
   <option  value="">Select state</option>
  <?php 
    $result=$wpdb->get_results("select * from states");
    foreach($result as $row) {
        $state_id=$row->state_id;
        $state_name=$row->state_name;
        echo '<option value='.$state_id.'>'.$state_name.'</option>';
    }
   ?>     
</select>

但是当我想编辑时，我如何显示第一个选择的州名。????

修改页面网址... whitecode.in/demo/plotsup_plot/new-property/?listing_edit=6795

这是我在function.php文件中的功能代码

   function getcity(){
      global $wpdb;
       if($_POST['state'])
            {
                $id=$_POST['state'];
                $property_id = $_GET['listing_edit']; 
                                     $district = get_post_meta($property_id, district, true);
                $result=$wpdb->get_results("SELECT * FROM districts WHERE state_id='$id'");
                //$wpdb->get_results($query);
                              foreach($result as $row) {
                                                             $city_name   = $row->district_name;
                             $city_id     = $row->district_id;
                        ?>
     <option value="<?php echo $city_id; ?>" <?php if($district == $city_id){ echo 
  'selected="selected"';} ?>><?php echo $city_name; ?></option>
  <?php      
                              //echo '<option value="'.$city_id.'">'.$city_name.'</option>';


            }
   } 
   }

Answer 1

根据之前评论中的讨论，我猜该属性是您网站中的自定义帖子类型。因此，如果这是情景，那么这应该是解决方案。

<select name="state"  id="state"  class="select-submit2">
<option  value="">Select state</option>
<?php 
$property_id = $_GET['listing_edit']; //the PROPERTY_ID should be replaced with the original id might be ina get variable or any process by which you are using for the edit page.
$property_state = get_post_meta($property_id, META_KEY_STATE, true); //META_KEY_STATE is the meta_key name you use to store the value of the state in the postmeta table
$result=$wpdb->get_results("select * from states");
foreach($result as $row) {
    $state_id=$row->state_id;
    $state_name=$row->state_name;
    ?>
    <option value="<?php echo $state_id; ?>" <?php if($property_state == $state_id){ echo 'selected="selected"';} ?>><?php echo $state_name; ?></option>
    <?php
}
?>     
</select>

尝试此操作，如果您遇到任何问题，请与我们联系。

Answer 2

如果这是你要找的东西？

<select name="state"  id="state"  class="select-submit2">
   <option  value="">Select state</option>
   <?php 
    $result=$wpdb->get_results("select * from states");
    foreach($result as $row) {
         $state_id=$row->state_id;
         $state_name=$row->state_name;
         if($state_id == SELECTED_STATE_ID)
             echo '<option value='.$state_id.' selected>'.$state_name.'</option>';
         else
             echo '<option value='.$state_id.'>'.$state_name.'</option>';
     }
   ?>     
</select>

只需将SELECTED_STATE_ID替换为包含所选状态的变量

如何在wordpress中的下拉列表中显示所选值

2 个答案: