Question

我编写了一个程序，该程序从Klients生日的sql值中获取并保存在i=[]中，但是我从datetime.datetime(xxxxx)那里获取了很多数据，我只需要年份，因为我希望获得我们客户的平均水平。

代码：

import psycopg2

from datetime import *


#verbindung mit dem server bzw. db
try:
    conn = psycopg2.connect("dbname='test' user='test' host='192.168.1.10' password='test'")
    print ("Connectet **********")
except:
    print ("I am unable to connect to the database")
#verbindung mit der tabele
cur = conn.cursor()
try:
    cur.execute("""SELECT bday from client""")
    print ("yes")
except:
    print ("I can't drop our test database!")

result = cur.fetchall()


i = []
for row in result:
    i.append(row)

print (i)
print (len(result))

输出：

...
(datetime.datetime(1967, 11, 7, 0, 0),), (datetime.datetime(1976, 1, 1, 0, 0),), (datetime.datetime(1978, 5, 3, 0, 0),), (datetime.datetime(1969, 6, 6, 0, 0),), (datetime.datetime(1977, 12, 26, 0, 0),), (datetime.datetime(1944, 8, 22, 0, 0),), (datetime.datetime(1954, 5, 31, 0, 0),), (datetime.datetime(1942, 12, 4, 0, 0),), (datetime.datetime(1967, 10, 16, 0, 0),), (datetime.datetime(1941, 7, 1, 0, 0),), (datetime.datetime(1937, 5, 17, 0, 0),), (datetime.datetime(2006, 6, 5, 0, 0),), (datetime.datetime(1996, 11, 25, 0, 0),), (datetime.datetime(1975, 3, 2, 0, 0),), (datetime.datetime(2007, 10, 22, 0, 0),), (datetime.datetime(1996, 7, 26, 0, 0),), (datetime.datetime(1991, 4, 4, 0, 0),), (datetime.datetime(1938, 11, 29, 0, 0),), 
....

我需要解决此问题的帮助。

Answer 1

psycopg返回元组，如果您仅从数据库中获得1列，则它返回包含1个元素[("a", ), ("a", )...]的元组列表

由于您获得日期时间，因此只能像row[0].year那样获得年份

for row in result:
    i.append(row[0].year)
print(i)

或类似的

i = [row[0].year for row in result]
print(i)

Answer 2

这是您执行所需代码的版本：

import psycopg2
import datetime


# verbindung mit dem server bzw. db
try:
    conn = psycopg2.connect("dbname='test' user='test' host='192.168.1.10' password='test'")
    print("Connectet **********")
except:
    print("I am unable to connect to the database")
# verbindung mit der tabele
cur = conn.cursor()
try:
    cur.execute("""SELECT bday from client""")
    print("yes")
except:
    print("I can't drop our test database!")

rows = cur.fetchall()

i = [row.year for row, *_ in rows]

print(i)
print(len(i))

Python日期时间问题-仅年份

2 个答案: