Question

我是python的新手，我正在尝试运行命令行，但它失败并显示错误TypeError: xeger() takes exactly 2 arguments (3 given)。我无法理解原因，因为我只向xeger.xeger发送了两个参数：out = xeger.xeger(pattern, limit)。我了解self已隐藏但会自动添加到参数列表中，因此我已更改为xeger.xeger(self, pattern, limit)并将函数签名更改为def generate(self, limit, pattern)但我仍然遇到相同的错误。

import xeger
import click
@click.command()
@click.option('--limit', default=10, help='Number of chars to return')
@click.option('--pattern', prompt='pattern to match', help='regexp to match')

def generate(limit, pattern):
    """Simple program that receives a pattern and returns a matching string."""
    out = xeger.xeger(pattern, limit)
    click.echo('%s' % out)

if __name__ == '__main__':
   generate()

运行并出错：

python rand.py --limit 230
pattern to match: asda
Traceback (most recent call last):
  File "rand.py", line 13, in <module>
    generate()
  File "/usr/local/lib/python2.7/site-packages/click/core.py", line 716, in __call__
    return self.main(*args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/click/core.py", line 696, in main
    rv = self.invoke(ctx)
  File "/usr/local/lib/python2.7/site-packages/click/core.py", line 889, in invoke
    return ctx.invoke(self.callback, **ctx.params)
  File "/usr/local/lib/python2.7/site-packages/click/core.py", line 534, in invoke
    return callback(*args, **kwargs)
  File "rand.py", line 9, in generate
    out = xeger.xeger(pattern, limit)
TypeError: xeger() takes exactly 2 arguments (3 given)

Answer 1

查看他们的源代码，您可以做两件事：

不要发送limit参数 - 默认为10。
稍微不同地初始化它。

像这样的东西

from xeger import Xeger

_x = Xeger(limit=limit)
out = _x.xeger(pattern)

TypeError：xeger（）只需2个参数（给定3个）

1 个答案: