TypeError:序列项0:期望字符串,找到int

Question

我试图将字典中的数据插入数据库.我想迭代值并相应地格式化它们,具体取决于数据类型.这是我正在使用的代码片段:

def _db_inserts(dbinfo):
    try:
        rows = dbinfo['datarows']

        for row in rows:
            field_names = ",".join(["'{0}'".format(x) for x in row.keys()])
            value_list = row.values()

            for pos, value in enumerate(value_list):
                if isinstance(value, str):
                    value_list[pos] = "'{0}'".format(value)
                elif isinstance(value, datetime):
                    value_list[pos] = "'{0}'".format(value.strftime('%Y-%m-%d'))

            values = ",".join(value_list)

            sql = "INSERT INTO table_foobar ({0}) VALUES ({1})".format(field_names, values)

    except Exception as e:
        print 'BARFED with msg:',e

当我使用一些示例数据(见下文)运行算法时,我收到错误:

TypeError:序列项0:期望字符串,找到int

给出上述错误的value_list数据的示例是:

value_list = [377, -99999, -99999, 'f', -99999, -99999, -99999, 1108.0999999999999, 0, 'f', -99999, 0, 'f', -99999, 'f', -99999, 1108.0999999999999, -99999, 'f', -99999, 'f', -99999, 'f', 'f', 0, 1108.0999999999999, -99999, -99999, 'f', 'f', 'f', -99999, 'f', '1984-04-02', -99999, 'f', -99999, 'f', 1108.0999999999999] 

我究竟做错了什么？

Answer 1

string.join 连接字符串列表中的元素,而不是整数.

请改用此生成器表达式:

values = ','.join(str(v) for v in value_list)

Answer 2

虽然给定的列表理解/生成器表达式答案还可以,但我发现这更容易阅读和理解:

values = ','.join(map(str, value_list))

Answer 3

更换

values = ",".join(value_list)

同

values = ','.join([str(i) for i in value_list])

要么

values = ','.join(str(value_list)[1:-1])

Answer 4

cval和Priyank Patel的答案很棒.但是,要知道,一些值可能是unicode字符串,因此可能会导致str抛出一个UnicodeEncodeError错误.在这种情况下,请str通过该功能替换该功能unicode.

例如,假设字符串Libië(荷兰语为利比亚),在Python中表示为unicode字符串u'Libi\xeb':

print str(u'Libi\xeb')

抛出以下错误:

Traceback (most recent call last):
  File "/Users/tomasz/Python/MA-CIW-Scriptie/RecreateTweets.py", line 21, in <module>
    print str(u'Libi\xeb')
UnicodeEncodeError: 'ascii' codec can't encode character u'\xeb' in position 4: ordinal not in range(128)

但是,以下行不会引发错误:

print unicode(u'Libi\xeb') # prints Libië

所以,替换:

values = ','.join([str(i) for i in value_list])

通过

values = ','.join([unicode(i) for i in value_list])

为了安全起见.