VSCode、typescript、eslint和prettier->；空检查不起作用

我试着将TypeScriptLinting设置为使用prettier，而prettier工作时，我似乎丢失了基本的typescript错误检查。我怎样才能找回这个

^应该抱怨访问

foo

上的

.value

My

tsconfig.json

{
  "compilerOptions": {
    "outDir": "./dist/",
    "noImplicitAny": true,
    "module": "es6",
    "target": "es5",
    "jsx": "react",
    "allowJs": true,
    "sourceMap": true,
    "moduleResolution": "node",
    "baseUrl": "./src",
    "allowSyntheticDefaultImports": true,
    "paths": {
      "~/*": ["./*"]
    }
  },
}

{
  "extends": [
    "eslint:recommended",
    "plugin:@typescript-eslint/recommended",
    "prettier/@typescript-eslint",
    "plugin:prettier/recommended"
  ],
  "plugins": ["@typescript-eslint", "prettier"],
  "env": {
    "browser": true,
    "jasmine": true,
    "jest": true,
    "es6": true
  },
  "rules": {
    "prettier/prettier": [
      "error", 
      { "singleQuote": true, "semi": false }
    ],
    "@typescript-eslint/no-use-before-define": "off",
    "@typescript-eslint/no-explicit-any": "off",
    "@typescript-eslint/class-name-casing":  ["error", { "allowUnderscorePrefix": true }],
    "@typescript-eslint/interface-name-prefix": ["error", { "prefixWithI": "always", "allowUnderscorePrefix": true}],
    "@typescript-eslint/no-unused-vars": ["error", { "args": "none"}],
    "@typescript-eslint/no-non-null-assertion": "off",
    "semi":"off"
  },
  "parser": "@typescript-eslint/parser"
}

.eslintrc.json

{
  "compilerOptions": {
    "outDir": "./dist/",
    "noImplicitAny": true,
    "module": "es6",
    "target": "es5",
    "jsx": "react",
    "allowJs": true,
    "sourceMap": true,
    "moduleResolution": "node",
    "baseUrl": "./src",
    "allowSyntheticDefaultImports": true,
    "paths": {
      "~/*": ["./*"]
    }
  },
}

{
  "extends": [
    "eslint:recommended",
    "plugin:@typescript-eslint/recommended",
    "prettier/@typescript-eslint",
    "plugin:prettier/recommended"
  ],
  "plugins": ["@typescript-eslint", "prettier"],
  "env": {
    "browser": true,
    "jasmine": true,
    "jest": true,
    "es6": true
  },
  "rules": {
    "prettier/prettier": [
      "error", 
      { "singleQuote": true, "semi": false }
    ],
    "@typescript-eslint/no-use-before-define": "off",
    "@typescript-eslint/no-explicit-any": "off",
    "@typescript-eslint/class-name-casing":  ["error", { "allowUnderscorePrefix": true }],
    "@typescript-eslint/interface-name-prefix": ["error", { "prefixWithI": "always", "allowUnderscorePrefix": true}],
    "@typescript-eslint/no-unused-vars": ["error", { "args": "none"}],
    "@typescript-eslint/no-non-null-assertion": "off",
    "semi":"off"
  },
  "parser": "@typescript-eslint/parser"
}

---

为什么

foo.value

应该是错误的

foo

是一个包含值作为变量的对象，

logFoo（foo:{value:number | null}）

@PareshLomate

foo

的类型是

{value:number}null

，因此

foo

可以是

null

，如果

foo===null

我就不能访问

foo.value

你是说

tsc

strictNullChecks

（或者更全面的

strict

，包括

strictNullChecks

）吗？您的配置中没有这些选项，如果我启用上述选项，编译器将再次发出错误。请写一个答案，我将接受它。谢谢