Question

我反对如下:

{"resources":[{"srno":"1","name":"name1","projects":[{"prjname":"abc"}]},{"srno":"1","name":"name1","projects":[{"prjname":"def"}]},{"srno":"2","name":"name2","projects":[{"prjname":"abc"}]},{"srno":"4","name":"name4","projects":[{"prjname":"prq"}]},{"srno":"4","name":"name4","projects":[{"prjname":"stu"}]},{"srno":"4","name":"name4","projects":[{"prjname":"uvw"}]},{"srno":"5","name":"name5"}]}

How can I get resources grouped by id,name - each of that should contain list of projects? basically,flattened to structured i.e.

{"resources":[{"srno":"1","name":"name1","projects":[{"prjname":"abc"},{"prjname":"def"}]},{"srno":"2","name":"name2","projects":[{"prjname":"abc"}]},{"srno":"4","name":"name4","projects":[{"prjname":"prq"},{"prjname":"stu"},{"prjname":"uvw"}]},{"srno":"5","name":"name5"}]}

一个人可以分享“jsonata”的表述,从而产生同样的效果。

Answer 1

我可以用jsonata公式解决这一问题。 Demo @ https://try.jsonata.org/fjSXaUjQq

1 : $distinct(                          /*select distinct item from */               
2 :    $.resources.                     /*(     for each item in  resources */
3 :    (                                /*  ( */
4 :        $id :=  srno;                /*        set variable $id=item.srno */
5 :        {                            /*         { return key-pair */
6 :            "srno":srno,"name":name, /*            "srno"=item.srno,"name"=item.name, */
7 :            "projects":%.resources   /*            "projects"= ( from item-parent.resources */
8 :            [srno=$id]               /*            where resources.srno=$id ) */
9 :            .projects                /*            return projects */
10:         }                           /*        }  */
11:    )                                /*    ) */
12: )                                   /*)  */

友情链接