要回答在响应成功或失败之前阻止转换到新路由的原始问题：

因为您使用的是 redux thunk，所以动作创建者的成功或失败可能会触发重定向。我不知道你的具体动作/动作创建者是什么样的，但这样的事情可以工作：

import { browserHistory } from 'react-router'

export function loadInitialFormValues(formId) {
  return function(dispatch) {
    // hit the API with some function and return a promise:
    loadInitialValuesReturnPromise(formId)
      .then(response => {
        // If request is good update state with fetched data
        dispatch({ type: UPDATE_FORM_STATE, payload: response });

        // - redirect to the your form
        browserHistory.push('/myForm');
      })
      .catch(() => {
        // If request is bad...
        // do whatever you want here, or redirect
        browserHistory.push('/myForm')
      });
  }
}

跟进。在进入路由/在组件的 componentWillMount 上加载数据并显示微调器的常见模式：

来自关于异步操作的 redux 文档http://redux.js.org/docs/advanced/AsyncActions.html

• 通知减速器请求开始的动作。

减速器可以通过切换状态中的 isFetching 标志来处理此操作。这样 UI 就知道是时候显示微调器了。

• 通知减速器请求成功完成的动作。

reducer 可以通过将新数据合并到它们管理的状态并重置 isFetching 来处理此操作。UI 将隐藏微调器，并显示获取的数据。

• 通知减速器请求失败的操作。

减速器可以通过重置 isFetching 来处理这个动作。此外，一些 reducer 可能想要存储错误消息，以便 UI 可以显示它。

我按照下面的一般模式使用您的情况作为粗略的指导方针。你不必使用Promise

// action creator:
export function fetchFormData(formId) {
  return dispatch => {
    // an action to signal the beginning of your request
    // this is what eventually triggers the displaying of the spinner
    dispatch({ type: FETCH_FORM_DATA_REQUEST })

    // (axios is just a promise based HTTP library)
    axios.get(`/formdata/${formId}`)
      .then(formData => {
        // on successful fetch, update your state with the new form data
        // you can also turn these into their own action creators and dispatch the invoked function instead
        dispatch({ type: actions.FETCH_FORM_DATA_SUCCESS, payload: formData })
      })
      .catch(error => {
        // on error, do whatever is best for your use case
        dispatch({ type: actions.FETCH_FORM_DATA_ERROR, payload: error })
      })
  }
}

// reducer

const INITIAL_STATE = {
  formData: {},
  error: {},
  fetching: false
}

export default function(state = INITIAL_STATE, action) {
  switch(action.type) {
    case FETCH_FORM_DATA_REQUEST:
      // when dispatch the 'request' action, toggle fetching to true
      return Object.assign({}, state, { fetching: true })
    case FETCH_FORM_DATA_SUCCESS:
      return Object.assign({}, state, {
        fetching: false,
        formData: action.payload
      })
    case FETCH_FORM_DATA_ERROR:
      return Object.assign({}, state, {
        fetching: false,
        error: action.payload
      })
  }
}

// route can look something like this to access the formId in the URL if you want
// I use this URL param in the component below but you can access this ID anyway you want:
<Route path="/myForm/:formId" component={SomeForm} />

// form component
class SomeForm extends Component {
  componentWillMount() {
    // get formId from route params
    const formId = this.props.params.formId
    this.props.fetchFormData(formId)
  }

  // in render just check if the fetching process is happening to know when to display the spinner
  // this could also be abstracted out into another method and run like so: {this.showFormOrSpinner.call(this)}
  render() {
    return (
      <div className="some-form">
        {this.props.fetching ? 
          <img src="./assets/spinner.gif" alt="loading spinner" /> :
          <FormComponent formData={this.props.formData} />
        }
      </div>
    )
  }
}

function mapStateToProps(state) {
  return {
    fetching: state.form.fetching,
    formData: state.form.formData,
    error: state.form.error
  }
}

export default connect(mapStateToProps, { fetchFormData })(SomeForm)

首先，我想说的是，关于使用 react-router 的onEnter钩子获取数据是否是一种好的做法存在争论，不过这就是这样的事情：

您可以将 redux-store 传递给您的Router. 让以下成为您的 Root 组件，Router安装位置：

...
import routes from 'routes-location';

class Root extends React.Component {
  render() {
    const { store, history } = this.props;

    return (
      <Provider store={store}>
        <Router history={history}>
          { routes(store) }
        </Router>
      </Provider>
    );
  }
}
...

您的路线将类似于：

import ...
...

const fetchData = (store) => {
  return (nextState, transition, callback) => {
    const { dispatch, getState } = store;
    const { loaded } = getState().myCoolReduxStore;
    // loaded is a key from my store that I put true when data has loaded

    if (!loaded) {
      // no data, dispatch action to get it
      dispatch(getDataAction())
        .then((data) => {
          callback();
        })
        .catch((error) => {
          // maybe it failed because of 403 forbitten, we can use tranition to redirect.
          // what's in state will come as props to the component `/forbitten` will mount.
          transition({
            pathname: '/forbitten',
            state: { error: error }
          });
          callback();
        });
    } else {
      // we already have the data loaded, let router continue its transition to the route
      callback();
    }
  }
};

export default (store) => {
  return (
    <Route path="/" component={App}>
      <Route path="myPage" name="My Page" component={MyPage} onEnter={fetchData(store)} />
      <Route path="forbitten" name="403" component={PageForbitten} />
      <Route path="*" name="404" component={PageNotFound} />
    </Route>
  );
};

请注意，您的路由器文件正在以您的商店作为参数导出一个 thunk，如果您向上看，看看我们如何调用路由器，我们将商店对象传递给它。

可悲的是，在撰写react-router 文档时，我会返回 404，因此我无法将您指向(nextState, transition, callback)所描述的文档。但是，关于那些，从我的记忆中：

• nextState描述路由react-router将过渡到；

• transition执行功能可能是另一个过渡而不是 from nextState;

• callback 将触发您的路线转换完成。

另一个需要指出的想法是，使用 redux-thunk，您的调度操作可以返回一个Promise，请在此处的文档中查看。你可以在这里找到一个关于如何使用 redux-thunk 配置你的 redux store 的好例子。

为此，我制作了一个方便的钩子，适用于 react-router v5：

/*
 * Return truthy if you wish to block. Empty return or false will not block
 */
export const useBlock = func => {
    const { block, push, location } = useHistory()
    const lastLocation = useRef()

    const funcRef = useRef()
    funcRef.current = func

    useEffect(() => {
        if (location === lastLocation.current || !funcRef.current)
            return
        lastLocation.current = location

        const unblock = block((location, action) => {
            const doBlock = async () => {
                if (!(await funcRef.current(location, action))) {
                    unblock()
                    push(location)
                }
            }
            doBlock()
            return false
        })
    }, [location, block, push])
}

在你的组件中，像这样使用它：

const MyComponent = () => {
    useBlock(async location => await fetchShouldBlock(location))

    return <span>Hello</span>
}

在 async 函数返回之前不会发生导航；您可以通过返回来完全阻止导航true。