c# - How to find and update an item inside of a array (of an object in an array) {Accounts[x].Items[y].StatusCode = ...} -

i have simple nested array (below). goal represent these items in c# 'account' objects contain list of 'item' along other account info. can set structure time in mongodb have no idea how update statuscode on single item on single account.

{      "accounts":[         {            "name":"somename1",          "email":"email1@site.com",          "owneditems":[               {                  "itemid":55,                "statuscode":1             },             {                  "itemid":12,                "statuscode":2             }          ]       },       {            "name":"somename2",          "email":"email2@site.com",          "owneditems":[               {                  "itemid":100,                "statuscode":3             },             {                  "itemid":101,                "statuscode":4             }          ]       }    ] } 

my question how query find , update specific statuscode in setup?

essentially want following. of course, following code not update db.

public void update_statuscode(string accountname, int itemid, int newstatuscode) {     var foundaccount = collection.find(builders<accountsave>.filter.eq(acc => acc.name, accountname)).firstordefaultasync().result;     foreach (var item in foundaccount.owneditems)     {         if (item.itemid == itemid)             item.statuscode = newstatuscode;     } } 

edit: clarify, using 2.0 driver. also, yes know there no '_id's. being example, assume used properly.

you can join conditions

public async task updatestatus(string accountname, int itemid, int newstatuscode) {     var filter = builders<account>.filter.and(         builders<account>.filter.where(x => x.name == accountname),         builders<account>.filter.elemmatch(x => x.owneditems, x => x.itemid == itemid));      var update = builders<account>.update.set("owneditems.$.statuscode", newstatuscode);     await collection.updateoneasync(filter, update); }