Question

我在表单中有多个组件，如下所示：

<Form_element check_empty={true} required={true} rows="1" label="Question"</Form_element>

Form_element的定义是：

var Form_element = React.createClass({
  getInitialState: function () {
  return({valid:false});
  )},
  render:function()
  {
     return(<input required="" type={this.props.type}/>)
  }
})

如何在提交表单时检查多个Form_element是否无效。（在每个组件上使用refs是一个坏主意）

Answer 1

你可以用状态

来做

class FormElement extends Component {
   constructor(props){
     super(props);
   }
   render(){ 
      const { state,value,type } = this.props;
      let errorClass={color:'red'};
      let normalClass={color:'black'};
      return (
         <div>
         {     (state == 0)?
               <input style={errorClass} type={type} 
                      value={this.props.value}>:
               <input style={normalClass} type={type}
                      value={value}>   
         }

         </div>

      );
   }
}
class Parent extends Component {
   constructor(props){
     super(props);
     this.onCheck = this.onCheck.bind(this);
     this.state = {name:{value:'',status:0 },address:{ value:'',status:0} };
   }
   onCheck(){

    let ok = {};

    Object.keys(this.state).forEach(x=>{
       let element = this.state[x]; 
        // check if the status is alright
        if ( cond satisfy your validation ){
           ok[x] = this.state[x];
        }

    }); 
     // let's say that the checking was done we update the state to rerender

    this.setState(Object.assign({},this.state,ok);

   }

   render(){

    return (
      <div>
         <FormElement valid={this.state.name.status} type={'text'}></FormElement>
     <FormElement valid={this.state.address.status} type={'text'}></FormElement> 
     <button onPress={this.onClick}>submit</button> 
      </div>
     );
   } 
}

只需在此组件的主机容器中通过状态播放值，PS：仅通过setState（）

进行重新渲染

验证多个reactjs组件

1 个答案: