我正在尝试使用mechanize(v0.2.5)来处理页面上具有禁用图像作为表单元素之一的表单.当我尝试选择表单时,mechanize会引发一个AttributeError: control 'test' is disabled
地方test
是禁用控件的名称.例如,
br = mechanize.Browser(factory=mechanize.RobustFactory())
br.open("http://whatever...")
br.select_form(nr=0)
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导致此堆栈跟踪:
br.select_form(nr=0)
File "build\bdist.win32\egg\mechanize\_mechanize.py", line 499, in select_form
File "build\bdist.win32\egg\mechanize\_html.py", line 544, in __getattr__
File "build\bdist.win32\egg\mechanize\_html.py", line 557, in forms
File "build\bdist.win32\egg\mechanize\_html.py", line 237, in forms
File "build\bdist.win32\egg\mechanize\_form.py", line 844, in ParseResponseEx
File "build\bdist.win32\egg\mechanize\_form.py", line 1017, in _ParseFileEx
File "build\bdist.win32\egg\mechanize\_form.py", line 2735, in new_control
File "build\bdist.win32\egg\mechanize\_form.py", line 2336, in __init__
File "build\bdist.win32\egg\mechanize\_form.py", line 1221, in __setattr__
AttributeError: control 'test' is disabled
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检查机械化源代码,当有任何表单元素评估为a mechanize.SubmitControl
且没有预定义value
属性时,看起来总是会引发此错误.例如,以下表单会引发相同的错误:
<form action="http://whatever" method="POST">
<input name="test" type="submit" disabled="disabled" />
</form>
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我不确定这是否应该算作一个bug,但无论如何都有解决方法吗?例如,有没有办法可以在调用之前更改目标页面的HTML以启用禁用的控件br.select_form()
?
编辑
我已经提交了一个机械化补丁修复了这个问题.
可悲的是,它已经超过一年了,机械化上游仍未合并拉动请求.
同时你可以使用我编写的这个猴子补丁来解决这个bug,而无需手动安装补丁版本.希望在(if)0.2.6发布时解决此错误,因此补丁仅适用于0.2.5及更早版本.
def monkeypatch_mechanize():
"""Work-around for a mechanize 0.2.5 bug. See: https://github.com/jjlee/mechanize/pull/58"""
import mechanize
if mechanize.__version__ < (0, 2, 6):
from mechanize._form import SubmitControl, ScalarControl
def __init__(self, type, name, attrs, index=None):
ScalarControl.__init__(self, type, name, attrs, index)
# IE5 defaults SUBMIT value to "Submit Query"; Firebird 0.6 leaves it
# blank, Konqueror 3.1 defaults to "Submit". HTML spec. doesn't seem
# to define this.
if self.value is None:
if self.disabled:
self.disabled = False
self.value = ""
self.disabled = True
else:
self.value = ""
self.readonly = True
SubmitControl.__init__ = __init__
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这绝对是一个错误,向上游报告它,制作补丁,向上游提交它并同时使用修补版本是处理它的相当正确的方法。(感谢您选择这种方式。)
正如您所提到的,另一种方法是通过预处理源 HTML 来解决这个问题(如果您很着急,或者由于某种原因不能/不想使用修补版本,这可能很有用,但请注意,解决方法并不不帮助社区)。对于后处理,可以使用任何合适的方法 - 从 str.replace() 到使用BeautifulSoup或lxml进行 DOM 级处理。
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