python oracle 将结果放入字典并匹配给定的列表

Question

如何将 sql 结果中的给定列表匹配到字典中？这是我的代码：

import cx_Oracle

con = cx_Oracle.connect('fmstech','fmstech','fmsdev')
cur = con.cursor()
cur.execute("select * from num_prefix where network = 'GLOBE'")

globe = ['639988800000', '639066256904', '0150422 153023']
results = {}
for lines in cur:
    results[lines[0]]=lines[1]
    globe1 = globe[1][2:5]
    if globe1 in results:
        print '906 is exist'
    else:
        print '906 is not exist'

cur.close()
con.close()

但我得到了结果：906 is not exist

Answer 1

解决方法是这样的。使用 int()

globe1 = int(globe[1][2:5])

示例 1： with out int()

results = {817: 'globe', 906: 'globe'}
globe = ['639988800000', '639066256904', '0150422 153023']
globe1 = globe[1][2:5]
if globe1 in results:
        print '906 is exist'
else:
        print '906 is not exist'

o/p

906 is not exist

示例 2：使用 int()

results = {817: 'globe', 906: 'globe'}
globe = ['639988800000', '639066256904', '0150422 153023']
globe1 = int(globe[1][2:5])
if globe1 in results:
        print '906 is exist'
else:
        print '906 is not exist'

o/p

906 is exist

原因：

在字典中，这里的键是数字，而 globel 在您的代码中是字符串。所以将其转换为 NUMBER 并进行比较。

python oracle 将结果放入字典并匹配给定的列表

python oracle gets result into dictionary and matches the given list

python

oracle

dictionary

python-2.6