Question

function BasicDatePicker(){
  const [selectedDate, handleDateChange] = useState(new Date());

  return (
    <MuiPickersUtilsProvider utils={DateFnsUtils}>
      <DatePicker
        label="Basic example"
        value={selectedDate}
        onChange={handleDateChange}
        animateYearScrolling
      />
      </MuiPickersUtilsProvider>
  );
}


class GameList extends Component {
  state = {};

  render() {
    const { classes } = this.props;
    return (
      <div id="gameList">
        <BasicDatePicker />
      </div>
    );
  }
}


export default (GameList);

arrayList已恢复

ArrayList<String> single = new ArrayList<String>();
---------         
System.out.println(single)

期望的结果

[Document{{googContentType=realtime, packetsLost=0, id=ssrc_3841594688_recv, type=ssrc, timestamp=2019-07-11T07:18:43.413Z}}
,Document{{googContentType=realtime, packetsLost=0, id=ssrc_239620516_recv, type=ssrc, timestamp=2019-07-11T07:18:42.955Z}}]

Answer 1

不一定是最佳实践，但是您可以进行正则表达式替换来删除最终输出中不需要的键值对：

List<String> singleOut = single.stream()
    .map(x -> x.replaceAll("(?:googContentType|id|type)=\\S+\\s*", ""))
    .collect(Collectors.toList());

singleOut.forEach(System.out::println);

此打印：

Document{{packetsLost=0, timestamp=2019-07-11T07:18:43.413Z}}
Document{{packetsLost=0, timestamp=2019-07-11T07:18:42.955Z}}

这里最好的解决方案是返回生成此JSON风格对象的工具/代码，然后重新导出，只包含要显示的packetsLost和timestamp键。

如何通过键删除arrayList中的元素

1 个答案: