问题

如何创建一个解析器，可以解析可能包含行继续符=的{{1}}符号后面的值？例如

应解析为a = b\e,\ c,\ d。

我一般都是使用这个库和解析器组合器的新手。所以任何正确方向的指针都非常受欢迎。

我尝试了什么

测试

(KeyValuePair (Key, 'a'), (Value, 'b\e, c, d'))

生产

尝试一个

public class ConfigurationFileGrammerTest
{
    [Theory]
    [InlineData("x\\\n  y", @"x y")]
    public void ValueIsAnyStringMayContinuedAccrossLinesWithLineContinuation(
        string input, 
        string expectedKey)
    {
        var key = ConfigurationFileGrammer.Value.Parse(input);
        Assert.Equal(expectedKey, key);
    }
}

测试输出

    public static readonly Parser<string> Value =
        from leading in Parse.WhiteSpace.Many()
        from rest in Parse.AnyChar.Except(Parse.Char('\\')).Many()
            .Or(Parse.String("\\\n")
            .Then(chs => Parse.Return(chs))).Or(Parse.AnyChar.Except(Parse.LineEnd).Many())
        select new string(rest.ToArray()).TrimEnd();

尝试两个

Xunit.Sdk.EqualException: Assert.Equal() Failure
           ↓ (pos 1)
Expected: x y
Actual:   x\
           ↑ (pos 1)

测试输出

    public static readonly Parser<string> SingleLineValue =
        from leading in Parse.WhiteSpace.Many()
        from rest in Parse.AnyChar.Many().Where(chs => chs.Count() < 2 || !(string.Join(string.Empty, chs.Reverse().Take(2)).Equals("\\\n")))
        select new string(rest.ToArray()).TrimEnd();

    public static readonly Parser<string> ContinuedValueLines =
        from firsts in ContinuedValueLine.AtLeastOnce()
        from last in SingleLineValue
        select string.Join(" ", firsts) + " " + last;

    public static readonly Parser<string> Value = SingleLineValue.Once().XOr(ContinuedValueLines.Once()).Select(s => string.Join(" ", s));

Answer 1

您不得在输出中包含行继续。这是最后一次单元测试的唯一问题。解析延续\\\n时，必须从输出结果中删除它并返回空字符串。对不起，我不知道如何使用C＃sprache。也许有类似的东西：

Parse.String("\\\n").Then(chs => Parse.Return(''))

我使用combinatorix python库解决了这个问题。它是一个解析器组合库。 API使用函数而不是使用链式方法，但想法是一样的。

以下是包含评论的完整代码：

# `apply` return a parser that doesn't consume the input stream.  It
# applies a function (or lambda) to the output result of a parser.
# The following parser, will remove whitespace from the beginning
# and the end of what is parsed.
strip = apply(lambda x: x.strip())

# parse a single equal character
equal = char('=')

# parse the key part of a configuration line. Since the API is
# functional it reads "inside-out". Note, the use of the special
# `unless(predicate, parser)` parser. It is sometime missing from
# parser combinator libraries. What it does is use `parser` on the
# input stream if the `predicate` parser fails. It allows to execute
# under some conditions. It's similar in spirit to negation in prolog.
# It does parse *anything until an equal sign*, "joins" the characters
# into a string and strips any space starting or ending the string.
key = strip(join(one_or_more(unless(equal, anything))))

# parse a single carriage return character
eol = char('\n')

# returns a parser that return the empty string, this is a constant
# parser (aka. it always output the same thing).
return_empty_space = apply(lambda x: '')
# This will parse a full continuation (ie. including the space
# starting the new line.  It does parse *the continuation string then
# zero or more spaces* and return the empty string
continuation = return_empty_space(sequence(string('\\\n'), zero_or_more(char(' '))))

# `value` is the parser for the value part.  Unless the current char
# is a `eol` (aka. \n) it tries to parse a continuation, otherwise it
# parse anything. It does that at least once, ie. the value can not be
# empty. Then, it "joins" all the chars into a single string and
# "strip" from any space that start or end the value.
value = strip(join(one_or_more(unless(eol, either(continuation, anything)))))

# this basically, remove the element at index 1 and only keep the
# elements at 0 and 2 in the result. See below.
kv_apply = apply(lambda x: (x[0], x[2]))

# This is the final parser for a given kv pair. A kv pair is:
#
# - a key part (see key parser)
# - an equal part (see equal parser)
# - a value part (see value parser)
#
# Those are used to parse the input stream in sequence (one after the
# other). It will return three values: key, a '=' char and a value.
# `kv_apply` will only keep the key and value part.
kv = kv_apply(sequence(key, equal, value))


# This is sugar syntax, which turns the string into a stream of chars
# and execute `kv` parser on it.
parser = lambda string: combinatorix(string, kv)


input = 'a = b\\e,\\\n    c,\\\n    d'
assert parser(input) == ('a', 'b\\e,c,d')

如何处理＆＃39; line-continuation＆＃39;使用解析器组合器

问题

我尝试了什么

测试

生产

1 个答案: