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"}
]
}
答案 0 :(得分:0)
使用jsonb_array_elements
select j->>'from' as "from" from t
cross join jsonb_array_elements(s->'details') as j;