解析带转义单引号的字符串

问题描述 投票:2回答:1

我想解析一个单引号之间包含ASCII字符的字符串,并且该字符串可以包含连续的两个'换码的单引号。

''包含在单引号之间的字符串值->''等等...'

这应该导致:

包含在单引号之间的字符串值->',依此类推...

use nom::{
    bytes::complete::{tag, take_while},
    error::{ErrorKind, ParseError},
    sequence::delimited,
    IResult,
};

fn main() {
    let res = string_value::<(&str, ErrorKind)>("'abc''def'");

    assert_eq!(res, Ok(("", "abc\'def")));
}

pub fn is_ascii_char(chr: char) -> bool {
    chr.is_ascii()
}

fn string_value<'a, E: ParseError<&'a str>>(i: &'a str) -> IResult<&'a str, &'a str, E> {
    delimited(tag("'"), take_while(is_ascii_char), tag("'"))(i)
}

如何检测转义引号而不是字符串的结尾?

parsing rust nom
1个回答
0
投票

这很棘手,但是可以起作用:

//# nom = "5.0.1"
use nom::{
    bytes::complete::{escaped_transform, tag},
    character::complete::none_of,
    combinator::{recognize, map_parser},
    multi::{many0, separated_list},
    sequence::delimited,
    IResult,
};

fn main() {
    let (_, res) = parse_quoted("'abc''def'").unwrap();
    assert_eq!(res, "abc'def");
    let (_, res) = parse_quoted("'xy@$%!z'").unwrap();
    assert_eq!(res, "xy@$%!z");
    let (_, res) = parse_quoted("'single quotes -> '' and so on...'").unwrap();
    assert_eq!(res, "single quotes -> ' and so on...");
}

fn parse_quoted(input: &str) -> IResult<&str, String> {
    let seq = recognize(separated_list(tag("''"), many0(none_of("\'"))));
    let unquote = escaped_transform(none_of("\'"), '\'', tag("\'"));
    let res = delimited(tag("'"), map_parser(seq, unquote), tag("'"))(input)?;

    Ok(res)
}

一些解释:

  1. 解析器seq识别在双引号和其他任何东西之间交替的任何序列;
  2. [unquote将任何双引号转换为单引号;
  3. [map_parser然后将两者结合在一起以产生所需的结果。
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