我的案例是使用setState回调，而不是setState + useEffect

坏❌

  const closePopover = useCallback(
    () =>
      setOpen((prevOpen) => {
        prevOpen && onOpenChange(false);
        return false;
      }),
    [onOpenChange]
  );

好✅

  const closePopover = useCallback(() => setOpen(false), []);

  useEffect(() => onOpenChange(isOpen), [isOpen, onOpenChange]);

TL,博士; 对于我的案例，我修复警告的方法是将useState更改为useRef

react_devtools_backend.js:2574 Warning: Cannot update a component (`Index`) while rendering a different component (`Router.Consumer`). To locate the bad setState() call inside `Router.Consumer`, follow the stack trace as described in https://reactjs.org/link/setstate-in-render
    at Route (http://localhost:3000/main.bundle.js:126692:29)
    at Index (http://localhost:3000/main.bundle.js:144246:25)
    at Switch (http://localhost:3000/main.bundle.js:126894:29)
    at Suspense
    at App
    at AuthProvider (http://localhost:3000/main.bundle.js:144525:23)
    at ErrorBoundary (http://localhost:3000/main.bundle.js:21030:87)
    at Router (http://localhost:3000/main.bundle.js:126327:30)
    at BrowserRouter (http://localhost:3000/main.bundle.js:125948:35)
    at QueryClientProvider (http://localhost:3000/main.bundle.js:124450:21)

我所做的上下文的完整代码(从带有// OLD:的行更改为它们上面的行)。然而，这并不重要，只需尝试从useState更改为useRef!!

import { HOME_PATH, LOGIN_PATH } from '@/constants';
import { NotFoundComponent } from '@/routes';
import React from 'react';
import { Redirect, Route, RouteProps } from 'react-router-dom';
import { useAccess } from '@/access';
import { useAuthContext } from '@/contexts/AuthContext';
import { AccessLevel } from '@/models';

type Props = RouteProps & {
  component: Exclude<RouteProps['component'], undefined>;
  requireAccess: AccessLevel | undefined;
};

export const Index: React.FC<Props> = (props) => {
  const { component: Component, requireAccess, ...rest } = props;

  const { isLoading, isAuth } = useAuthContext();
  const access = useAccess();
  const mounted = React.useRef(false);
  // OLD: const [mounted, setMounted] = React.useState(false);

  return (
    <Route
      {...rest}
      render={(props) => {
        // If in indentifying authentication state as the page initially loads, render a blank page
        if (!mounted.current && isLoading) return null;
        // OLD: if (!mounted && isLoading) return null;

        // 1. Check Authentication is one step
        if (!isAuth && window.location.pathname !== LOGIN_PATH)
          return <Redirect to={LOGIN_PATH} />;
        if (isAuth && window.location.pathname === LOGIN_PATH)
          return <Redirect to={HOME_PATH} />;

        // 2. Authorization is another
        if (requireAccess && !access[requireAccess])
          return <NotFoundComponent />;

        mounted.current = true;
        // OLD: setMounted(true);
        return <Component {...props} />;
      }}
    />
  );
};

export default Index;

此警告从React V16.3.0开始引入。

如果您正在使用功能组件，您可以将setState调用包装到useEffect中。

不能工作的代码:

const HomePage = (props) => {
    
  props.setAuthenticated(true);

  const handleChange = (e) => {
    props.setSearchTerm(e.target.value.toLowerCase());
  };

  return (
    <div key={props.restInfo.storeId} className="container-fluid">
      <ProductList searchResults={props.searchResults} />
    </div>
  );
};

现在你可以把它改成:

const HomePage = (props) => {
  // trigger on component mount
  useEffect(() => {
    props.setAuthenticated(true);
  }, []);

  const handleChange = (e) => {
    props.setSearchTerm(e.target.value.toLowerCase());
  };

  return (
    <div key={props.restInfo.storeId} className="container-fluid">
      <ProductList searchResults={props.searchResults} />
    </div>
  );
};

我认为这很重要。 @Red-Baron在这篇文章中指出:

@machineghost:我认为你误解了这条信息警告的内容。

将更新父节点状态的回调传递给子节点并没有什么错。这一直都很好。

问题是当一个组件在另一个组件中排队更新时，而第一个组件正在呈现。

换句话说，不要这样做:

function SomeChildComponent(props) {
    props.updateSomething();
    return <div />
}

但这是可以的:

function SomeChildComponent(props) {
    // or make a callback click handler and call it in there
    return <button onClick={props.updateSomething}>Click Me</button>
}

而且，正如Dan多次指出的那样，在渲染时在同一个组件中排队更新也很好:

function SomeChildComponent(props) {
  const [number, setNumber] = useState(0);

  if(props.someValue > 10 && number < 5) {
    // queue an update while rendering, equivalent to getDerivedStateFromProps
    setNumber(42);
  }

  return <div>{number}</div>
}

使用上面的一些答案，我摆脱了以下错误:

from

if (value === "newest") {
  dispatch(sortArticlesNewest());
} else {
  dispatch(sortArticlesOldest());
}

这段代码位于我的组件顶层

to

    const SelectSorting = () => {
  const dispatch = useAppDispatch();

  const {value, onChange} = useSelect();

  useEffect(() => {
    if (value === "newest") {
      dispatch(sortArticlesNewest());
    } else {
      dispatch(sortArticlesOldest());
    }
  }, [dispatch, value]);

我也有同样的问题。我设置了一些存储函数的状态，像这样:

// my state definition
const [onConfirm, setOnConfirm] = useState<() => void>();

// then I used this piece of code to update the state
function show(onConfirm: () => void) {
    setOnConfirm(onConfirm);
}

问题来自setOnConfirm。在React中，setState可以接受新值，也可以接受返回新值的函数。在这种情况下，React希望通过调用onConfirm来获得新的状态，这是不正确的。

改成这个解决了我的问题:

setOnConfirm(() => onConfirm);