Question

我创建了州和城市的依赖性下拉。这是财产提交表格。当用户编辑表单时，我想显示所选的州和城市。我设法以某种方式显示选定的状态。我写了get_city函数function.php并通过ajax调用它。我有两个不同的表，即州和城市。所以问题是我不明白如何显示选定的城市？这是我的下拉代码

<select name="state" id="state" class="select-submit2">
   <option value="">Select state</option>

   <?php 
   $property_id = $_GET['listing_edit']; 
   $property_state = get_post_meta($property_id, state, true);
   $result=$wpdb->get_results("select * from states");
   //$wpdb->get_results($query);
   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 $state_name; ?></option>
<?php 
      //echo '<option value='.$state_id.'>'.$state_name.'</option>';
   }
?>          
</select>
<input type="hidden" name="edit_id" id="edit_id" value="<?php echo $edit_id;?>">
<select name="district" id="district" required class="select-submit2">
   <option value="">Select District</option>
</select>

在上面的代码中我显示了选中的状态。我在google上搜索，发现我必须将post id传递给function.php中的函数。这是我在function.php文件中的get_city函数的代码

function getcity() {

global $wpdb;

   if($_POST['state']) {
      $id = $_POST['state'];
      global $edit_id;
      $property_id = $edit_id; 
      $district = get_post_meta($property_id, 'district', true);
      $result=$wpdb->get_results("SELECT * FROM districts WHERE state_id='$id'");

      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>';                 
      }
   }
}
add_action("wp_ajax_nopriv_getcity", "getcity");
add_action("wp_ajax_getcity", "getcity");

这是我的js代码......

 jQuery(document).ready(function($) {

  $('#state').on('change', function() {

   var state  = $('#state').val();
   var edit_id = $('#edit_id').val();

   $.ajax
   ({
    type : "POST",
    url  : "http://whitecode.in/demo/plotsup_plot/wp-admin/admin-ajax.php",
    data :{'action' : 'getcity', 'state' : state , 'edit_id' : edit_id },
    success: function(html) {

      $("#district").removeAttr("disabled");
      $("#district").html(html);
     }

    });
   });

 <input type="hidden" name="edit_id" id="edit_id" value="<?php echo $edit_id;?>">

Answer 1

您必须将上述代码放在一个函数中。让函数名称为myFunction()。如果要将post id传递给函数，请调用post id作为参数的函数，即myFunction($postid)其中$postid是post id。请不要忘记从post循环调用函数，以便函数调用每个postid。

Answer 2

我个人为每个州加载a，其中style =＆＃34; display：none;＆＃34;，之后，当用户选择状态时，相应的更改样式=＆＃34; display：none;＆＃34; to style =＆＃34; display：inline;＆＃34;

Answer 3

试试这个，让我知道它是否有效。更改是在两行全局$ edit_id上进行的;已删除，$ property_id行已更改。请正确应用更改并告知我们是否正常。

<?php
function getcity() {

global $wpdb;

if($_POST['state']) {
  $id = $_POST['state'];
  $property_id = $_POST['edit_id']; 
  $district = get_post_meta($property_id, 'district', true);
  $result=$wpdb->get_results("SELECT * FROM districts WHERE state_id='$id'");

  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>';                 
  }
}
}
add_action("wp_ajax_nopriv_getcity", "getcity");
add_action("wp_ajax_getcity", "getcity");

如何在用户编辑任何表单时显示所选的下拉值？

3 个答案: