为什么ast.literal_eval（）似乎忽略声明的变量？ [英] Why does ast.literal_eval() seem to ignore declared variables?

问题描述

我听说ast.literal_eval比eval（）安全得多，但是在更改代码时，出现格式错误的字符串/节点错误。

I have heard that ast.literal_eval is much safer than eval(), but while changing my code, I am getting 'malformed string/node' errors.

例如：

bar = False
incorrect = {"foo":bar}
correct = {"foo":"bar"}

ast.literal_eval(incorrect) 

返回错误，但

ast.literal_eval(correct) 

返回期望的{ foo： bar}

returns the expected {"foo":"bar"}

为什么第一次评估不返回{ foo：False}

Why doesn't the first evaluation return {"foo":False}

推荐答案

因为这并不意味着这样做。从文档中：

Because it is not meant to do that. From the documentation:

安全地评估表达式节点或包含Python表达式的Unicode或Latin-1编码的
字符串。提供的字符串或节点可能
仅包含以下Python文字结构：字符串，
数字，元组，列表，字典，布尔值和无。

Safely evaluate an expression node or a Unicode or Latin-1 encoded string containing a Python expression. The string or node provided may only consist of the following Python literal structures: strings, numbers, tuples, lists, dicts, booleans, and None.

ast.literal_eval 和 eval 的意思是将字符串表示形式转换为将python代码转换为...有效的python代码。

ast.literal_eval and eval are meant to turn a string representation of python code into ...valid python code.

这将不起作用：

>>> ast.literal_eval({"foo": "bar"})
Traceback (most recent call last):
  File "<input>", line 1, in <module>
  File "/usr/lib/python2.7/ast.py", line 80, in literal_eval
    return _convert(node_or_string)
  File "/usr/lib/python2.7/ast.py", line 79, in _convert
    raise ValueError('malformed string')
ValueError: malformed string <-- Telltale right here.

这是因为您已经具有要评估的有效python结构。

That is because you already have a valid python structure which you are attempting to evaluate.

如果将引号括起来，它将创建一个字典：

If you put quotes around the whole thing instead, it will create a dictionary:

>>> ast.literal_eval('{"foo": "bar"}')
{'foo': 'bar'}

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