JavaScript对象中obj.key形式删除操作是否仅支持字符串键？非字符串键删除方式解析

Great question! Let's unpack what's happening here and answer your core questions clearly.

First, a quick foundational note: All object keys (except Symbols) are implicitly converted to strings in JavaScript. So when you define const a = {a: 1, 0: 2}, the key 0 is actually stored as the string "0" under the hood.

Why delete a.0 throws an error #

The dot notation (obj.key) has strict syntax rules: the part after the dot must be a valid JavaScript identifier. Valid identifiers can't start with a number, can't contain hyphens/spaces, and can't be reserved keywords. Since 0 starts with a number, it's not a valid identifier—so a.0 is a syntax error before the delete operation even runs.

Answering your questions #

1. Does obj.key-style deletion only work for string-type keys?
   More precisely: it only works for keys that are valid JavaScript identifiers (which are treated as strings). Not all string keys qualify—for example, keys like "0", "user-name", or "123abc" can't be accessed with dot notation, even though they're strings. You'll need square brackets for those cases.

2. Do non-string keys require obj[key] syntax for deletion?
   Let's break this down by key type:

   • Number keys: As we saw, numbers are converted to strings, but you can't use dot notation (since numbers aren't valid identifiers). You must use obj[key] (either obj[0] or obj["0"]—both work, since 0 gets coerced to "0").
   • Symbol keys: Symbols are a special key type that aren't converted to strings. Dot notation doesn't support Symbols, so you have to use obj[mySymbol] to delete them.
   • Other non-string types: Any other non-string (like a boolean or object) will be converted to a string when used as a key, but again, if the resulting string isn't a valid identifier, you'll need square brackets.

Example to confirm #

For your original object:

const a = {a: 1, 0: 2};
delete a.a; // Works, since "a" is a valid identifier
delete a["0"]; // Works, accesses the string key "0"
delete a[0]; // Also works, since 0 is coerced to "0"
delete a.0; // Syntax error—invalid identifier

内容的提问来源于stack exchange，提问作者David542