Kau*_*ana 8 python mongodb pymongo
我正试图在MongoDB中找到一条记录,并_id从结果中过滤.
这是我的代码:
#app.py
@app.route('/login', methods = ['GET', 'POST'])
def login():
if request.method == "POST":
password = request.form.get('password')
email = request.form.get('email')
db = get_db()
data = db.author.find_one({'email' : email, 'password' : password})
print(data)
return 'data'
else:
return render_template('login.html')
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输出:
{'password': '123123', 'name': '<my_name>', 'email': '<my_email>', '_id': ObjectId('<an_object_id_string>')}
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如何_id从输出中过滤字段?
sty*_*ane 13
您需要使用投影指定要返回的字段.
data = db.author.find_one({'email' : email, 'password' : password}, {'_id': 1})
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小智 5
这是避免id的最好方法,
data = db.author.find_one({'email' : email, 'password' : password},{"password":1, "email":1, "name":1,"_id": False})
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现在你得到了 ANSWER "{'password': '123123', 'name': 'prakash', 'email': 'prakashprabhu48@gmail.com'}"(没有 id)