代码之家 › 专栏 › 技术社区 › Dave

在pyparsing中使用QuotedString

pyparsing parsing python

4

Dave · 技术社区 · 11 年前

我在理解如何构建 pyparsing 解析器。步骤是:1)通过组合ParserElement的子类来构建解析器,2)使用解析器来解析字符串。

以下示例工作正常:

from pyparsing import Word, Literal, alphas, alphanums, delimitedList, QuotedString

name = Word(alphas+"_", alphanums+"_")
field = name
fieldlist = delimitedList(field)
doc = Literal('<Begin>') + fieldlist + Literal('**End**')

dstring = '<Begin>abc,de34,f_o_o**End**'
print(doc.parseString(dstring))

产生预期的令牌序列:

['<Begin>', 'abc', 'de34', 'f_o_o', '**End**']

但是(例如),类QuotedString不接受ParserElement作为参数,因此它不能用于构建解析器。我希望在上面的示例中使用它,例如:

name = Word(alphas+"_", alphanums+"_")
field = QuotedString(name)     ### Wrong: doesn't allow "name" as an argument
fieldlist = delimitedList(field)

要解析以下形式的文档:

dstring = '<Begin>"abc", "de34", "f_o_o"**End**'

但是,既然它不能以这种方式使用,那么在构造一个引用字符串列表的解析器时包含QuotedString的正确语法是什么?

==========编辑============

请参阅下面的答案。。。

2 回复 | 直到 11 年前

1

4

Dave 11 年前

QuotedString不能用于此任务。但是OR函数可以实现相同的效果-允许不同形式的引号,同时保留分析引号中包含的字符串的有效性的能力。以下代码执行此操作:

from pyparsing import Word, Literal, alphas, alphanums, delimitedList
from pyparsing import Group, QuotedString, ParseException, Suppress

name = Word(alphas+"_", alphanums+"_")
field = Suppress('"') + name + Suppress('"') ^ \    # double quote
        Suppress("'") + name + Suppress("'") ^ \    # single quote
        Suppress("<") + name + Suppress(">") ^ \    # html tag
        Suppress("{{")+ name + Suppress("}}")       # django template variable
fieldlist = Group(delimitedList(field))
doc = Literal('<Begin>') + fieldlist + Literal('**End**')

dstring = [
    '<Begin>"abc","de34","f_o_o"**End**',      # Good
    '<Begin><abc>,{{de34}},\'f_o_o\'**End**',  # Good
    '<Begin>"abc",\'de34","f_o_o\'**End**',    # Bad - mismatched quotes
    '<Begin>"abc","de34","f_o#o"**End**',      # Bad - invalid identifier
]

for ds in dstring:
    print(ds)
    try:
        print('  ', doc.parseString(ds))
    except ParseException as err:
        print(" "*(err.column-1) + "^")
        print(err)

这会产生所需的输出,接受两个好的测试字符串,拒绝两个坏的测试字符串:

<Begin>"abc","de34","f_o_o"**End**
   ['<Begin>', ['abc', 'de34', 'f_o_o'], '**End**']
<Begin><abc>,{{de34}},'f_o_o'**End**
   ['<Begin>', ['abc', 'de34', 'f_o_o'], '**End**']
<Begin>"abc",'de34","f_o_o'**End**
            ^
Expected "**End**" (at char 12), (line:1, col:13)
<Begin>"abc","de34","f_o#o"**End**
                   ^
Expected "**End**" (at char 19), (line:1, col:20)

感谢Paul的所有帮助和制作如此酷的包装。

2

1

PaulMcG 11 年前

我想您只是对如何使用QuotedString有点困惑。传递给QuotedString的参数为不需要在引号内的字符串-它是用作引号字符的字符。通过这种方式,您可以定义使用“*”作为引号、使用“=”作为引号或使用“<”的带引号字符串和“>”开始和结束引号字符。在您的示例中,只需对字段使用以下定义:

field = QuotedString('"')

此外,不要害怕使用python的内置help()方法来访问类、模块、方法等的文档字符串。

编辑:

QuotedString('X') 做不作语法分析 "X" ,它解析 X some characters inside matching characters X .

以下是您的完整(工作)示例程序:

from pyparsing import QuotedString, delimitedList, Group

dstring = '<Begin>"abc", "de34", "f_o_o"**End**'
field = QuotedString('"')
parser = "<Begin>" + Group(delimitedList(field)) + "**End**"

print(parser.parseString(dstring))

这是我的指纹:

['<Begin>', ['abc', 'de34', 'f_o_o'], '**End**']

如果复制/粘贴此示例并运行它后出现异常, 请发布完整的异常。

更多示例:

starQuoteString = QuotedString('*')
eqQuoteString = QuotedString('=')
tildeQuoteString = QuotedString('~')
angleQuoteString = QuotedString('<', endQuoteChar='>')

fullSample = starQuoteString + eqQuoteString + tildeQuoteString + angleQuoteString

print fullSample.parseString("""
    *a string quoted with stars*
    =a very long quoted string, contained within equal signs=
    ~not a very long string at all~<another quoted string on the same line>
    """)

打印:

['a string quoted with stars', 'a very long quoted string, contained within equal signs', 'not a very long string at all', 'another quoted string on the same line']

您甚至不限于单个字符。你可以使用 QuotedString('**') 解析你的结束语 **End** ,但这也可以接受 **The End** 或 **Finis** 或 **That's all folks!** .