2 回答

TA貢獻1871條經驗 獲得超13個贊
Project.find({ uid: req.user._id },{updated_date:0,created_date:0})
.then((project) => {
res.status(200).send({
status: "success",
data: {project},
message: "successful"
});
})
您可以選擇隱藏某些字段,例如{updated_date:0,created_date:0}這不會顯示updated_date和created_date。但如果你使用 1 它會顯示

TA貢獻1911條經驗 獲得超7個贊
您可以簡單地使用delete運算符,例如:
let response = {
"status": "success",
"data": {
"project": [
{
"_id": "5e8baea9fe729e4d78855332",
"uid": "5e89c1bd2566a6360cf78097",
"project_name": "Test 00",
"updated_date": "2020-04-06T22:35:21.555Z",
"created_date": "2020-04-06T22:35:21.555Z",
"__v": 0
},
{
"_id": "5e8bb4f5fadc384eb73c6d59",
"uid": "5e89c1bd2566a6360cf78097",
"project_name": "Test 11",
"updated_date": "2020-04-06T23:02:13.498Z",
"created_date": "2020-04-06T23:02:13.498Z",
"__v": 0
}
]
},
"message": "successful"
}
response.data.project.map((p) => {
delete p.created_date;
delete p.updated_date;
});
console.log(response.data.project);
JavaScriptdelete運算符從對象中刪除屬性;如果不再持有對同一屬性的引用,它最終會自動釋放。
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