python TypeError:参数1必须是字符串或只读字符缓冲区,而不是None

时间:2015-01-02 15:04:34

标签: python regex python-2.7 typeerror

这是我的代码:     def main():

    if len(sys.argv) != 3:

        fname = "OP.tex"
        dot_ind = fname.index(".") + 1
        ofname = fname[:dot_ind] + "1." + fname[dot_ind:]

    else:

        fname = sys.argv[1]
        ofname = sys.argv[2]
    print (fname, "+",ofname)
    writeout(ofname, replacement(readin(fname)))
def replacement(content):
   pattern = re.compile(r'(?<=\\\\\[-16pt]\n)([\s\S]*?)(?=\\\\\n\\thinhline)')
   re.findall(pattern, content)

if __name__ == "__main__":
    main()

我收到以下错误:

File "test.py", line 30, in main
writeout(ofname, replacement(readin(fname)))
File "/home/utilities.py", line 138, in writeout
out.write(content)
TypeError: argument 1 must be string or read-only character buffer, not None

writeout是另一个文件中的函数,它告诉python输出内容的位置如下:

def writeout(filename, content, append=False):
    """
    Writes content to file filename.
    """

    mode = "w"

    #append to the file instead of overwriting
    if append:
        mode = "a"

    #write content
    with open(filename, mode) as out:
        out.write(content)

fname和ofname(输入和输出文件名)是正确的,那么为什么当程序是字符串时程序会说参数为None?

非常感谢你。

1 个答案:

答案 0 :(得分:6)

您的replacement函数必须返回一个字符串,目前它返回None。所以,改变声明

re.findall(pattern, content)

类似

return ' '.join(re.findall(pattern, content))

请注意,您不能只返回re.findall的结果,因为这是一个列表,而不是字符串;你必须以某种方式列出该结果的清单。

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