Question

如果我有一个像这样的对象数组

[{name: 'james', name: 'john'}]，我知道john的索引，并且想要更改john的值。

person = person.map((p, i)=>i===index?({...p, name: 'changed john'})):p) this.setState({person})

但是如果数组像这样怎么办？ ['james', 'john']

Answer 1

但是如果数组像这样怎么办？ ['james'，'john']

由于更新状态的正确方法是使用[oslo_messaging] topic = octavia_prov # Topic for octavia's events sent to a queue event_stream_topic = neutron_lbaas_event方法，因此，像setState这样直接改变状态的行为将不起作用。我认为我们照常使用this.state.list[i]="Changed John"。像这样：

Array.prototype.map()

class App extends React.Component {
  constructor(props) {
    super(props);
    this.state = {
      list: ['james', 'john']
    };
  }
  handleDelete = i => {
    const newList = this.state.list.filter((li, idx) => idx !== i);
    this.setState({ list: newList });
  };
  handleChange = i => {
    const newList = this.state.list.map(
      (li, idx) => (idx === i ? 'Changed ' + li : li)
    );
    this.setState({ list: newList });
  };
  render() {
    return (
      <div>
        {this.state.list.map((e, i) => (
          <div key={i}>
            <p>{e}</p>
            <button onClick={() => this.handleDelete(i)}>Delete</button>
            <button onClick={() => this.handleChange(i)}>Change</button>
          </div>
        ))}
      </div>
    );
  }
}
ReactDOM.render(<App />, document.getElementById('root'));

Answer 2

您想要这样的东西吗？

person = person.map((p, i) => p === "john" ? "changed john" : p);

Answer 3

这很容易。这里有两种情况的示例：

const test = [{name: 'james', name: 'john'}];

// Here we're accessing the 1st element and its 'name' property
console.log(test[0].name);

// Here we're accessing the 2nd element.
// JavaScript arrays are zero-indexed: the first element of an array is at index 0

const test2 = ['james', 'john'];

console.log(test2[1])

我建议访问 MDN docs，以探索其他（所有）示例。

Answer 4

person = person.map（（p，i）=> p ===“ john”？“ changed john”：p）;

使用地图更新没有对象的数组

4 个答案: