ReactJS:使用函数作为道具返回<;的值;td>;给予<;组件/>;错误
我有以下功能:ReactJS:使用函数作为道具返回<;的值;td>;给予<;组件/>;错误,reactjs,state,Reactjs,State,我有以下功能: budgetList() { return this.state.projects.map(currentproject=> { return <Budget project={currentproject} filter= {this.filterIssues} key={currentproject._id} stop={this.stopPropagation} modal={this.state.modal} toggle={this.t
budgetList() {
return this.state.projects.map(currentproject=> {
return <Budget project={currentproject} filter= {this.filterIssues} key={currentproject._id} stop={this.stopPropagation} modal={this.state.modal} toggle={this.toggle}/>;
})
}
现在我使用函数为预算元素中的每个元素检索一个值,如下所示
const Budget = props => (
<tr onClick={() => props.toggle(props.project._id)}>
<td>£{((props.project.proposedBudget / 4) *1000) }</td>
<td>{() => props.filter(props.project.project, "APP" + 74)}</td>
<td>{() => props.filter(props.project.project, "APP" + 75)}</td>
</tr>
)
const预算=道具=>(
props.toggle(props.project.\u id)}>
((props.project.proposedBudget/4)*1000)
{()=>props.filter(props.project.project,“APP”+74)}
{()=>props.filter(props.project.project,“APP”+75)}
)
问题是每一个
函数作为子函数无效。这可能会发生,如果你
返回组件,而不是从渲染返回。或者你
用于调用此函数而不是返回它
不太清楚我做错了什么。函数本身在预算组件之外工作正常。如果要呈现筛选函数返回的
和
,应该执行它<代码>{props.filter(props.project.project,“APP”+74)}有效,谢谢:)
const Budget = props => (
<tr onClick={() => props.toggle(props.project._id)}>
<td>£{((props.project.proposedBudget / 4) *1000) }</td>
<td>{() => props.filter(props.project.project, "APP" + 74)}</td>
<td>{() => props.filter(props.project.project, "APP" + 75)}</td>
</tr>
)