未绑定本地错误。为什么会发生这种情况？答案

【问题标题】：UnboundLocalError. Why has this occurred?未绑定本地错误。为什么会发生这种情况？
【发布时间】：2014-04-06 13:57:39
【问题描述】：

c.execute("SELECT * FROM Student, Behaviour")
data = c.fetchall() #'data' is a tuple and cannot be altered
l = []
for i in data: #for individual records in the whole database do:
    record = str(i)
    record = record.replace("u'","")
    record = record.replace("'", "")
    record = record.replace('"', '')
    record = record.replace("(","")
    record = record.replace(")", "")
    Formattedrecord = record.replace(","," -")
l.append(Formattedrecord)

这会返回错误：UnboundLocalError: local variable 'Formattedrecord' referenced before assignment 但是我不明白为什么，有人可以帮忙吗？

【问题讨论】：

缩进最后一行作为循环的一部分。你得到了UnboundLocalError，因为data 是空的，但如果它不是空的，你只需将last 条目附加到l。
你为什么在这里把record变成str()，然后把所有的引号都换掉？也许您想添加' - '.join(i) 而不是。

标签： python list for-loop sqlite

【解决方案1】：

注意缩进：

c.execute("SELECT * FROM Student, Behaviour")
data = c.fetchall() #'data' is a tuple and cannot be altered
l = []
for i in data: #for individual records in the whole database do:
    record = str(i)
    record = record.replace("u'","")
    record = record.replace("'", "")
    record = record.replace('"', '')
    record = record.replace("(","")
    record = record.replace(")", "")
    Formattedrecord = record.replace(","," -")
    l.append(Formattedrecord) # this should be part of the for loop

【讨论】：

OP 可能不想想使用str(i);从外观上看，Formattedrecord = ' - '.join(i) 会更合适。