遵循如何將 ab[cd][e].f[g[h[ij]]] 清楚地表示為對象樹？，您將如何編寫一個算法來從表達式生成 JS AST a.b[c.d][e].f[g[h[i.j]]]？我正在嘗試編寫一個解析器來從這個表達式生成某種對象結構（理想情況下比 JS AST 更直觀MemberExpression，因此是另一個問題）。我想看看算法是如何構建 JavaScriptMemberExpression樹的。
目前我有這種算法來生成某種樹（但目前似乎不正確）：
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]
}期望的結果是這樣的（或者如果您願意創建一個更好、更直觀的數據結構）：
{
"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
}
我掙扎的原因（部分）是我不喜歡這種MemberExpression樹狀結構，它是落後的感覺並且不是很直觀。因此，如果您可以構建一個更簡單、更直接的數據結構，那將是理想的（這是另一個問題），但如果不能，那麼僅構建一個算法就可以讓我繼續前進。
就我個人而言，我更願意嘗試生成這種結構，因為我覺得它更直觀：
{
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'
}
]
}
]
}
]
}
]
}
但其中任何一個都適合我（或兩者）。
如果我們使用第二個，我的下一個問題將是如何將該數據結構轉換為MemberExpression樹/數據結構 :) 但我會先嘗試自己做。所以在這個問題中構建 MemberExpression 可能更好，然後我可以解決這個問題。
將字符串分成第一級的對象和屬性組，例如
[
"a",
"b",
"[c.d]",
"[e]",
"f",
"[g[h[i.j]]]"
]
獲取對象
computed為true並從周圍的括號中剝離該屬性。type: "MemberExpression" ,object 與對象（2.），property與調用主函數getAST（1.）的結果，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; }本文收集自互联网，转载请注明来源。
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