我正在为api使用laravel数据库分页。它输出如下:
{
"current_page": 1,
"data": [
{
"id": 1,
"name": "Default"
},
{
"id": 2,
"name": "Default without Sidebar"
}
],
"first_page_url": "http://example.com/api/layouts?a=1",
"from": 1,
"last_page": 1,
"last_page_url": "http://example.com/api/layouts?a=1",
"next_page_url": null,
"path": "http://example.com/api/layouts",
"per_page": 10,
"prev_page_url": null,
"to": 2,
"total": 2
}
我不想要或不需要first_page_url
,last_page_url
等,所以应该只是:
{
"current_page": 1,
"data": [
{
"id": 1,
"name": "Default"
},
{
"id": 2,
"name": "Default without Sidebar"
}
],
"from": 1,
"last_page": 1,
"to": 2,
"total": 2
}
那么,有没有办法,格式化输出json,以便我可以删除或隐藏一些我不需要的额外信息。
答案 0 :(得分:0)
您可以这样做:
$except = ['first_page_url', 'last_page_url'];
return array_except(<YourModel>::paginate(...)->toArray(), $except);