database - 在 python 中比较时迭代两个数据库数据集

Question

下面提到的代码用于比较相同或不同数据库中的两个表。我无法检索到想要的结果以获得不匹配的记录。

我的问题:我无法在 table1 和 table2 中打印唯一不匹配的记录，因为我发现很难逐行迭代。目前，即使是匹配的记录也会打印为不匹配的记录。

import psycopg2
conn_string = "host='localhost' dbname='dvdrental' user='postgres' password='jai'"
db1 = psycopg2.connect(conn_string)
db2= psycopg2.connect(conn_string)
cursor1=db1.cursor()
cursor2=db2.cursor()
cursor1.execute("select * from public.actor order by 1")
results1 = cursor1.fetchall()
cursor2.execute("select * from public.actor order by 1")
results2 = cursor2.fetchall()
count1 =  len(results1)
count2 =  len(results2)
# print count1
# print count2
# print results1
# print results2
# print results1[0]
# print results2[0]
for i in range(0,count1):
    for j in range(0,count2):
        if (results1[i] == results2[j]):
            print "found"
        else:
            print "not found",results1

Answer 1

我试过类似下面的代码。请让我知道您的反馈

import psycopg2
conn_string = "host='localhost' dbname='dvdrental' user='postgres' password='jai'"
db1 = psycopg2.connect(conn_string)
db2= psycopg2.connect(conn_string)
cursor1=db1.cursor()
cursor2=db2.cursor()
cursor1.execute("select * from public.actor except select * from public.actor_1")
results1 = cursor1.fetchall()
cursor2.execute("select * from public.actor_1 except select * from public.actor")
results2 = cursor2.fetchall()
count1 =  len(results1)
count2 =  len(results2)
# print count1
# print count2
# print results1
# print results2
print results1
print results2