如何写一个分析构JavaScript AST成员表达的一个。b[c。d][e]。f[g[h[i。j]]]?

0

的问题

下沿线的 如何清楚的表示。b[c。d][e]。f[g[h[i。j]]]作为一个对象树?, 你会怎么写的一个算法产生的,JS AST从表达式 a.b[c.d][e].f[g[h[i.j]]]? 我想写一个分析程序产生某种形式的对象结构,从这一表(理想的情况更加直观的比JS AST MemberExpression 一,因此,其它问题)。 我想看看如何算法的工作构建JavaScript MemberExpression 树。

目前,我有这样的算法产生 的某些种类的 树木(但它似乎是不正确的前):

const patterns = [
  [/^[a-z][a-z0-9]*(?:-[a-z0-9]+)*/, 'name'],
  [/^\[/, 'open'],
  [/^\]/, 'close'],
  [/^\./, 'stem']
]

console.log(parsePath('a.b[c.d][e].f[g[h[i.j]]]'))

function parsePath(str) {
  let node
  let nest = []
  let result = nest
  let stack = [nest]
  while (str.length) {
    nest = stack[stack.length - 1]
    p:
    for (let pattern of patterns) {
      let match = str.match(pattern[0])
      if (match) {
        if (pattern[1] === 'name') {
          node = {
            form: `term`,
            name: match[0],
            link: []
          }
          nest.push(node)
        } else if (pattern[1] === 'stem') {
          stack.push(node.link)
        } else if (pattern[1] === 'open') {
          node = {
            form: 'read',
            link: []
          }
          nest.push(node)
          stack.push(node.link)
        } else if (pattern[1] === 'close') {
          stack.pop()
        }

        str = str.substr(match[0].length)
        break p
      }
    }
  }
  return result[0]
}

所希望的结果是这个(或更好、更直观数据结构,如果你是那么倾向于创建一个):

{
  "type": "MemberExpression",
  "object": {
    "type": "MemberExpression",
    "object": {
      "type": "MemberExpression",
      "object": {
        "type": "MemberExpression",
        "object": {
          "type": "MemberExpression",
          "object": {
            "type": "Identifier",
            "name": "a"
          },
          "property": {
            "type": "Identifier",
            "name": "b"
          },
          "computed": false
        },
        "property": {
          "type": "MemberExpression",
          "object": {
            "type": "Identifier",
            "name": "c"
          },
          "property": {
            "type": "Identifier",
            "name": "d"
          },
          "computed": false
        },
        "computed": true
      },
      "property": {
        "type": "Identifier",
        "name": "e"
      },
      "computed": true
    },
    "property": {
      "type": "Identifier",
      "name": "f"
    },
    "computed": false
  },
  "property": {
    "type": "MemberExpression",
    "object": {
      "type": "Identifier",
      "name": "g"
    },
    "property": {
      "type": "MemberExpression",
      "object": {
        "type": "Identifier",
        "name": "h"
      },
      "property": {
        "type": "MemberExpression",
        "object": {
          "type": "Identifier",
          "name": "i"
        },
        "property": {
          "type": "Identifier",
          "name": "j"
        },
        "computed": false
      },
      "computed": true
    },
    "computed": true
  },
  "computed": true
}

原因我苦苦挣扎的(部分)是我不喜欢这个 MemberExpression 树形结构,这是落后的感觉并不是非常直观的。 所以如果你可以构建一个更简单的更简单的数据结构,将是理想(这是其他问题),但如果不再只是一个算法建造这会让我去的。

就个人而言,我宁愿尝试产生这种结构,因为我找到它更加直观:

{
  type: 'site',
  site: [
    {
      type: 'term',
      term: 'a'
    },
    {
      type: 'term',
      term: 'b'
    },
    {
      type: 'sink',
      sink: [
        {
          type: 'term',
          term: 'c'
        },
        {
          type: 'term',
          term: 'd'
        }
      ]
    },
    {
      type: 'sink',
      sink: [
        {
          type: 'term',
          term: 'e'
        }
      ]
    },
    {
      type: 'term',
      term: 'f'
    },
    {
      type: 'sink',
      sink: [
        {
          type: 'term',
          term: 'g'
        },
        {
          type: 'sink',
          sink: [
            {
              type: 'term',
              term: 'h'
            },
            {
              type: 'sink',
              sink: [
                {
                  type: 'term',
                  term: 'i'
                },
                {
                  type: 'term',
                  term: 'j'
                }
              ]
            }
          ]
        }
      ]
    }
  ]
}

但无论哪一个对我的作品(或两者)。

如果我们有第二个,我的下一个问题将如何转换,数据结构进入 MemberExpression 树的数据结构:)但是我会试着做我自己的第一个。 所以它可能是更好地构造的成员在这个问题,那么我可以工作。

1

最好的答案

1
  1. 独立的串入群体的对象和特性的第一级,喜欢

    [
        "a",
        "b",
        "[c.d]",
        "[e]",
        "f",
        "[g[h[i.j]]]"
    ]
    
  2. 得到对象

    1. 采取的最后一项作为财产。
    2. 如果检查了酒店开始与支架上,然后设置 computedtrue 并带的酒店从周边的括号。
    3. 返回的对象
      • type: "MemberExpression" ,
      • object 与对象(2.),
      • property 与结果的主要功能 getAST (1.),
      • computed.

function getAST(string) {

    function getObject(parts) {
        if (parts.length === 1) return { type: "Identifier", name: parts[0] };

        let property = parts.pop(),
            computed = false;

        if (property.startsWith('[')) {
            computed = true;
            property = property.slice(1, -1);
        }

        return {
            type: "MemberExpression",
            object: getObject(parts),
            property: getAST(property),
            computed
        };
    }

    let i = 0,
        dot,
        bracket,
        parts = [];

    while (i < string.length) {
        dot = string.indexOf('.', i);
        bracket = string.indexOf('[', i);

        if (dot !== -1 && (bracket === -1 || dot < bracket)) {
            const temp = string.slice(i, dot);
            if (temp) parts.push(temp);
            i = dot + 1;
            continue;
        }

        if (bracket !== -1 && (dot === -1 || bracket < dot)) {
            const temp = string.slice(i, bracket);
            if (temp) parts.push(temp);
            i = bracket;

            let open = 1,
                j = i;

            while (++j < string.length) {
                if (string[j] === '[') open++;
                if (string[j] === ']') open--;
                if (!open) break;
            }

            j++;
            parts.push(string.slice(i, j));

            i = j;
            continue;
        }
        parts.push(string.slice(i));
        break;
    }

    return getObject(parts);
}

console.log(getAST('a.b[c.d][e].f[g[h[i.j]]]'));
.as-console-wrapper { max-height: 100% !important; top: 0; }

2021-11-24 07:30:42

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