我想在我的组件中触发一个 Action 。这是一个演示组件,不需要Redux感知。路由器的实现是使用react-router-redux完成的。
main.js:
let store = createStore(rootReducer)
const history = syncHistoryWithStore(hashHistory, store)
class RenderClass extends Component {
render() {
return (
<Provider store={store}>
<Router history={history}>
<Route path="/" component={MainLayout}>
<Route name="employees" path="/employees" url="http://localhost:8080/" component={EmployeeTable}></Route>
<Route name="personalInformation" path="/personalInformation/:employeeId" url={URI} component={PersonalInformation} />
.....
</Route>
<Router>
</Provider>
);
}
}
App.jsx:
import * as Actions from './action-creator';
const MainLayout = React.createClass({
render: function() {
const { dispatch, list } = this.props;
let actions = bindActionCreators(Actions, dispatch);
console.log(actions);
return (
<div className="wrapper">
<Header actions={actions} list={list} params={this.props.params} />
{React.cloneElement(this.props.children, this.props)}
</div>
)
}
});
function select(state) {
return {
list: state.listingReducer
}
}
export default connect(select)(MainLayout);
header.js:
define(
[
'react',
'jquery',
'appMin'
],
function (React, $) {
var Link = require('reactRouter').Link;
var Header = React.createClass({
handleClick: function () {
//Action called here
this.props.actions.listEmployees();
},
render: function () {
return (
<ul className="sidebar-menu">
<li>
<Link to={'/employees'} onClick={this.handleClick}><i className="fa fa-home"></i>Employees</Link>
</li>
</ul>
);
}
});
return Header;
}
)
employee.js:
define(
[
'react',
'jquery',
'appMin'
],
function (React, $) {
var EmployeeTable = React.createClass({
render: function () {
if (this.props.list != undefined) {
var listItems = this.props.list.map(function (listItem) {
return (
<tr key={listItem.id}>
<td> {listItem.name}</td>
<td><Link to={'/personalInformation/' + employee.id} onClick={this.props.actions.displayPersonalInformation(employee.id)}>Personal Information</Link></td>
......
</tr>
);
}, this);
})
}
return (
<table>
<tbody>
{listItems}
</tbody>
</table>
);
}
})
}
)
action-creator.js:
export function listEmployees() {
return {
type: types.LIST_EMPLOYEES
};
}
export function displayPersonalInformation() {
return {
type: types.DISPLAY_PERSONAL_INFORMATION
};
}
reducer.js:
function addEmployee(state, action) {
switch (action.type) {
case types.LIST_EMPLOYEES:
return {"id":"1", "name":"Stackoverflow"}
default:
return state
}
}
function listingReducer(state = [], action) {
switch (action.type) {
case types.LIST_EMPLOYEES:
return [
...state,
addEmployee(undefined, action)
]
case types.DISPLAY_PERSONAL_INFORMATION:
return // Gets personal Information
default:
return state;
}
}
const rootReducer = combineReducers({
listingReducer
})
export default rootReducer
Prop 不会包含 Action ,因为我没有将其绑定(bind)到 Prop 。我尝试在 App.js 中编写mapStateToProps和mapDispatchToProps,如下所示:
function mapStateToProps(state) {
return {
list: state.list
}
}
function mapDispatchToProps(dispatch) {
return { actions: bindActionCreators({ actionCreators }, dispatch) }
}
导出默认的connect(mapStateToProps,mapDispatchToProps)(MainLayout)
我正在调度不是函数错误。错误语句实际上听起来像是stackoverflow中的重复问题,但我也想知道为什么我使用mapDispatchToProps的目的也是正确的。提前致谢。
最佳答案
因此,据我了解,您正在使用export default connect(mapStateToProps, mapDispatchToProps)(MainLayout)
并尝试在dispatch
中调用MainLayout
。
因此,如果您看到react-redux文档,则可以通过以下方式调用connect:
connect()(//Component)
->调度将作为 Prop connect(mapStateToProps)(//Component)
->调度将作为 Prop connect(mapStateToProps,mapDispatchToProps)(//Component)
-> 调度不会作为 Prop 传递到组件 在方案3中,行为是这样的,因为
mapDispatchToProps
已经可以使用调度功能。因此,您可以将函数绑定(bind)到mapDispatchToProps
中进行分派(dispatch),然后将此函数传递给您的组件。示例
您要分派(dispatch)的函数是
dispatchMe()
和dispatchMe2()
,它们调用一个称为dispatchAction
的操作。因此,您的mapDispatchToProps
将类似于:-function mapDispatchToProps(dispatch){
return {
dispatchMe : () => {
dispatch(dispatchAction())
},
dispatchMe2 : () => {
dispatch(dispatchAction2())
}
}
}
现在,您可以在组件中传递
dispatchMe
并在需要时调用它。您可以阅读here以获得更多有关此的信息
关于reactjs - 如何将 Action 传递给Redux中的组件,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/40733822/