【发布时间】:2018-09-14 03:28:43
【问题描述】:
我正在从名为blocked_sites 的数据库表中获取值。如果blocked_sites 表中第0 个属性的值出现在文件items.csv 的第19 或第26 个字段中,那么将从csv 文件中排除该csv 行。我正在为此编写代码并收到此错误:
$ python csv_dupli_prev.py
Traceback (most recent call last):
File "csv_dupli_prev.py", line 48, in <module>
found = re.search(row[0], row1[19])
File "/home/debarati/anaconda3/lib/python3.6/re.py", line 182, in search
return _compile(pattern, flags).search(string)
File "/home/debarati/anaconda3/lib/python3.6/re.py", line 300, in _compile
raise TypeError("first argument must be string or compiled pattern")
TypeError: first argument must be string or compiled pattern
代码如下:
connection = pymysql.connect (host = "localhost", user = "root", passwd = "......", db = "city_details")
cursor = connection.cursor ()
csv_file = csv.reader(open("items.csv", "r"))
newrows = []
cursor.execute ("select * from blocked_sites")
data4 = cursor.fetchall ()
for row in data4:
for row1 in csv_file:
str1 = row1[19]
str2 = row1[26]
found = re.search(row[0], str1)
found1 = re.search(row[0], str2)
if found==None and found1==None and row1 not in newrows:
newrows.append(row1)
writer = csv.writer(open("items.csv", "w"))
writer.writerows(newrows)
【问题讨论】:
-
仅从错误消息来看(感谢包含整个错误消息 - 不是每个人都这样做),
row[0]不是字符串或编译模式。在调试器中查看它或使用print语句查看它是什么。 -
@ArndtJonasson :我发现了这个错误并修复了它。检查我的答案。