将前置项目反应到列表中

Question

我正在尝试将项目添加到列表中：

addPersonHandler = () => {
    const newPerson = {
          id: "new",
          edit: true,        
          name: "",
          dni: "",
          onDelete: this.discardHandler
        };

    // I want to prepend the new person to the people list.
    this.setState({addingPerson: true, people: {[newPerson].concat(this.state.people)});
}

但它总是呈现最后！

<ul>
    People.map((p, i) => {
      return <li key={p.id}>
        <Person 
          id={p.id} 
          name={p.name} 
          dni={p.dni} 
          onDelete={p.id=="new"? this.discardHandler: this.deleteHandler}
          edit={p.edit}         
          />
      </li>
    });    
</ul>

我真的不知道为什么如果我将它添加到列表的开头，它将呈现最后...

Answer 1

您可以在原始数组上使用spread，并删除包装新数组的{}

this.setState({addingPerson: true, people: [newPerson, ...this.state.people]);

Answer 2

尝试这样，在一个新的数组变量中取状态并将新对象推入其中，最后更新状态。

  addPersonHandler = () => {
      const newPerson = {
        id: "new",
        edit: true,
        name: "",
        dni: "",
        onDelete: this.discardHandler
      };

      let pplArr = this.state.people;
      pplArr.push(newPerson);
      this.setState({ addingPerson: true, people: pplArr });
    };

Answer 3

考虑the unshift() method，您可以使用它来将一个或多个元素添加到people数组中。

请记住，unshift()方法会改变它所调用的数组：

addPersonHandler = () => {

    const newPerson = {
          id: "new",
          edit: true,        
          name: "",
          dni: "",
          onDelete: this.discardHandler
        };


       // Get people array from current state
       const people = this.state.people;

       // Prepends the newPerson to the start of people array
       people.unshift(newPerson);

       // Return the updated state to your component for re-render
       this.setState({ addingPerson : true, people : people });
}

Answer 4

<ul>
          <Person 
          id={p.id} 
          name={p.name} 
          dni={p.dni} 
          onDelete={p.id=="new"? this.discardHandler: this.deleteHandler}
          edit={p.edit}         
          />
    People.map((p, i) => {
      return <li key={p.id}>

      </li>
    });    
</ul>