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I'm working on a templating engine where some of the syntax could be like:

{{ somevar|filter }}

In place of somevar could be an arbitrary "expression", which is to say, either a variable name like somevar, or a nested filter expression (like {{ somevar|filter|anotherfilter }}). I'm trying to parse this using Rust's nom parser combinator library, but failing to get it to work so far.

Here's the parser I've come up with so far:

#[macro_use]
extern crate nom;

use std::str;

#[derive(Debug)]
pub enum Expr<'a> {
    Var(&'a [u8]),
    Filter(&'a str, Box<Expr<'a>>),
}

#[derive(Debug)]
pub enum Node<'a> {
    Lit(&'a [u8]),
    Expr(Expr<'a>),
}

named!(expr_var<Expr>, dbg_dmp!(map!(nom::alphanumeric, Expr::Var)));

named!(expr_filter<Expr>,
    dbg_dmp!(do_parse!(
         val: any_expr >>
         tag_s!("|") >>
         name: map_res!(nom::alphanumeric, str::from_utf8) >>
         (Expr::Filter(name, Box::new(val)))
    ))
);

named!(any_expr<Expr>, dbg_dmp!(ws!(
    alt_complete!(
        expr_filter |
        expr_var  
    ))));

named!(expr_node<Node>, dbg_dmp!(map!(
    delimited!(tag_s!("{{"), any_expr, tag_s!("}}")),
    Node::Expr)));

named!(parse_template< Vec<Node> >, many1!(expr_node));

With a playground. The current version panics through a stack overflow. I can fix this by reversing the expr_var | expr_filter order in any_expr, but then I'm back to basically the same error as before.

4

2 回答 2

0

我不能说我在挖掘你的问题：没有应该解析的文本示例，你也没有描述你在构建解析器时遇到的问题。

不过，也许下面的例子会有所帮助。一个工作递归解析器：

#[macro_use]
extern crate nom;

use nom::alphanumeric;

type Variable = String;
type Filter = String;

named! (plain_expression (&str) -> (Variable, Filter), do_parse! (
    tag_s! ("{{") >>
    variable: alphanumeric >>
    tag_s! ("|") >>
    filter: alphanumeric >>
    tag_s! ("}}") >>
    ((variable.into(), filter.into()))));

#[derive(Debug)]
enum Expression {
    Plain(Variable, Filter),
    Recursive(Box<Expression>, Filter),
}

named! (recursive_expression (&str) -> Expression,
  alt_complete! (
    map! (plain_expression, |(v, f)| Expression::Plain (v, f)) |
    do_parse! (
      tag_s! ("{{") >>
      sub: recursive_expression >>
      tag_s! ("|") >>
      filter: alphanumeric >>
      tag_s! ("}}") >>
      (Expression::Recursive (Box::new (sub), filter.into())))));

fn main() {
    let plain = "{{var|fil}}";
    let recursive = "{{{{{{var1|fil1}}|fil2}}|fil3}}";
    // Prints: Done("", ("var", "fil")).
    println!("{:?}", plain_expression(plain));
    // Prints: Done("", Recursive(Recursive(Plain("var1", "fil1"), "fil2"), "fil3")).
    println!("{:?}", recursive_expression(recursive));
}

（操场）。

于 2017-01-04T11:56:59.380 回答
0

我通过编写自己的解析器函数来修复它：

named!(expr_var<Expr>, map!(nom::alphanumeric, Expr::Var));

fn expr_filtered(i: &[u8]) -> IResult<&[u8], Expr> {
    let (mut left, mut expr) = match expr_var(i) {
        IResult::Error(err) => { return IResult::Error(err); },
        IResult::Incomplete(needed) => { return IResult::Incomplete(needed); },
        IResult::Done(left, res) => (left, res),
    };
    while left[0] == b'|' {
        match nom::alphanumeric(&left[1..]) {
            IResult::Error(err) => {
                return IResult::Error(err);
            },
            IResult::Incomplete(needed) => {
                return IResult::Incomplete(needed);
            },
            IResult::Done(new_left, res) => {
                left = new_left;
                expr = Expr::Filter(str::from_utf8(res).unwrap(), Box::new(expr));
            },
        };
    }
    return IResult::Done(left, expr);
}

named!(expr_node<Node>, map!(
    delimited!(tag_s!("{{"), ws!(expr_filtered), tag_s!("}}")),
Node::Expr));

可能有一些更好的方法可以用 nom 宏做同样的事情，但至少我得到了一些工作。

于 2017-01-06T13:40:04.113 回答