有人知道我如何在不“替换”它包含的所有信息的情况下将对象添加到名为worker_info的数组中吗? (我正在学习平均堆栈)我有这段代码,但是它的作用是替换所有信息并擦除我已经拥有的其他数据。有了我得到的代码,总结一下我想要的是将worker_info数组添加到worker对象中而不删除其余部分,如果有人可以帮助我,我将不胜感激,任何建议都将非常有用,谢谢
我尝试过
User.findOneAndUpdate(serviceId, { "$push": { services: { worker_info: worker } } }, { new: true }, (err, us) => {
res.status(200).send(us);
});
我期望这样的事情
{
"services": [
{
"worker_info": [],
"_id": "5ce485a3288efe1834d4d71c",
"user_address": "dbarva252",
"description_service": "descripcion servicio",
"skill_type": "wasting time",
"amount_money": 0
},
{
"worker_info": [{
"_id": "5ce4867e288efe1834d4d71e",
"name": "Trabajador",
"surname": "arreglador",
"identification": "123457669",
"birthdate": "1 febrero 1999",
"skills": "time waster fixer",
"image": null,
"__v": 0
}],
"_id": "5ce485c5288efe1834d4d71d",
"user_address": "esta es una direccion",
"description_service": "descripcion servicio 2",
"skill_type": "wasting time part 2",
"amount_money": 0
}
],
"_id": "5ce48569288efe1834d4d71a",
"user": "usuario123",
"name": "mauricio",
"surname": "barva",
"identification": "123123",
"birthdate": "6 febrero 2000",
"city": "city",
"address": "tv 1d norte",
"cellphone": "123",
"image": null,
"status": false,
"__v": 0
}
但是我明白了
{
"services": [
{
"worker_info": [],
"_id": "5ce485a3288efe1834d4d71c",
"user_address": "dbarva252",
"description_service": "descripcion servicio",
"skill_type": "wasting time",
"amount_money": 0
},
{
"worker_info": {
"_id": "5ce4867e288efe1834d4d71e",
"name": "Trabajador",
"surname": "arreglador",
"identification": "123457669",
"birthdate": "1 febrero 1999",
"skills": "time waster fixer",
"image": null,
"__v": 0
}
}
],
"_id": "5ce48569288efe1834d4d71a",
"user": "usuario123",
"name": "mauricio",
"surname": "barva",
"identification": "123123",
"birthdate": "6 febrero 2000",
"city": "city",
"address": "tv 1d norte",
"cellphone": "123",
"image": null,
"status": false,
"__v": 0
}
如您所见,它将替换所有信息。