在PostgreSQL中对json数组元素进行迭代

Question

我在postgresql数据库的列中有两个JSON数据行，看起来像这样。

<script type="text/ng-template">

  $(".dropdown-menu li a").click(function () {
    $(this).parents(".dropdown").find('.btn').html($(this).text() + ' <span class="caret"></span>');
    $(this).parents(".dropdown").find('.btn').val($(this).data('value'));
  });
</script>

<ng-template #content let-modal>

   <div class="modal-body">
    <form>
      <div class="dropdown">
        <button class="btn btn-default dropdown-toggle" type="button" data-toggle="dropdown" aria-haspopup="true" aria-expanded="true">
          Dropdown
          <span class="caret"></span>
        </button>
        <ul class="dropdown-menu" aria-labelledby="dropdownMenu1">
          <li><a href="#" data-value="action">Action</a></li>
          <li><a href="#" data-value="another action">Another action</a></li>
          <li><a href="#" data-value="something else here">Something else here</a></li>
          <li><a href="#" data-value="separated link">Separated link</a></li>
        </ul>
      </div>
    </form>
  </div>

  
</ng-template>

我想遍历详细信息，并使用postgresql中的查询仅获取“ from”键值。 我想要它

{
  "details":[{"to":"0:00:00","from":"00:00:12"}]
}

{
  "details":[ 
             {"to":"13:01:11","from":"13:00:12"}, 
             {"to":"00:00:12","from":"13:02:11"}
            ]
}

Answer 1

使用jsonb_array_elements

select j->>'from' as "from" from t 
cross join jsonb_array_elements(s->'details') as j;

Demo