Question

每当我收到一个新的即将到来的道具时，我想改变当地的状态。为此，我使用lifecycle库中的Recompose HOC。但它并不像我想象的那么简单。我要么永远设置值，要么在使用回调时发生堆栈溢出。

import { connect } from 'react-redux';
import { withState, withProps, lifecycle, compose } from 'recompose';
import { selectors, testDelete, testSubscribe } from 'ducks/entities/tests';
import { Tests } from 'components/tests/all-tests';

export default compose(
  connect(selectors.tests, {
    onTestDelete: testDelete,
    onTestSubscribe: testSubscribe,
  }),
  withState('isDeleteModalOpen', 'setDeleteModalShow', false),
  withState('idToDelete', 'setIdToDelete', 0),
  withProps(
    ({
      tests,
      idToDelete,
      setIdToDelete,
      setDeleteModalShow,
      onTestDelete,
      onTestSubscribe,
    }) => ({
      tests: tests.map((t) =>
        t.merge({
          onDeleteModalShow: () => {
            setDeleteModalShow(true);
            setIdToDelete(t.get('id'));
          },
          onSubscribe: () => onTestSubscribe(t.get('id')),
        }),
      ),
      onDeleteModalHide: () => setDeleteModalShow(false),
      onDelete: () => onTestDelete(idToDelete),
    }),
  ),
  lifecycle({
    componentWillReceiveProps({ setDeleteModalShow }) {
      setDeleteModalShow(false); // Not workding "Maximum update depth exceeded"

      this.setState({
        isDeleteModalOpen: false,
      }); // now working
    },
  }),
)(Tests);

Answer 1

有点老问题，但答案很简单。

如果您设置的值已经是您想要的值，请不要更改状态。在您的代码中，您可以更改lifecycle增强器以检查isDeleteModalOpen是否已经为假，如下所示：

lifecycle({
  componentWillReceiveProps({ setDeleteModalShow, isDeleteModalOpen }) {
    if (isDeleteModalOpen) setDeleteModalShow(false);
  },
}),

这样，你的componentWillReceiveProps函数会弹出两次，第二次不会执行任何操作。

请注意，此组件上的任何道具更改都会将isDeleteModalOpen值设置为false。

groupWillReceiveProps生命周期内的setState重构HOC

1 个答案: