mysql - React/Redux - 将项目存储在状态中并将其发送到数据库

Question

我正在构建一个应用程序，该应用程序的一部分是它允许人们收藏某些项目。他们只需单击 SVG 心形图标，它就会变成红色，并以状态存储到“收藏夹”，并发送到数据库 (MySQL)。如果他们再次单击该图标，它会变回灰色，将其从状态中删除，并再次运行数据库提取以删除该项目。

我一直在尝试将其构建到我的 Action 创建器中，但我无法让它们都工作(也就是说，我无法将其发送到状态和数据库)。然而，我确实让它完美地适合国家作品。我把它放在下面:

export const toggleFavorite = name => (dispatch, getState) => {
  const { searchResults, favorites } = getState()
  const currentSelection = filter(
    selection => selection.name === name,
    searchResults
  )
  const newFavorite = contains(currentSelection, favorites)
    ? reject(equals(currentSelection), favorites)
    : append(currentSelection, favorites)
  dispatch({
    type: SET_FAVORITE,
    payload: newFavorite
  })
}

所以，我的问题是:在哪里以及如何将我的提取数据放到后端以允许插入和删除？这是我最后一次失败的尝试:

export const toggleFavorite = name => async (dispatch, getState) => {
  const { searchResults, favorites } = getState()
  const currentSelection = filter(
    selection => selection.name === name,
    searchResults
  )
  const newFavorite = contains(currentSelection, favorites)
    ? reject(equals(currentSelection), favorites)
      await fetch(`http://localhost:5000/deletefavorite?user_id=1&selection_id=${currentSelection}`)
    : append(currentSelection, favorites)
      await fetch(`http://localhost:5000/addfavorite?user_id=1&selection_id=${currentSelection}`)

  dispatch({
    type: SET_FAVORITE,
    payload: newFavorite
  })
}

提前致谢!

Answer 1

您应该用经典的 if/else 替换三元条件:

if (contains(currentSelection, favorites)) {
    const newFavorite = reject(equals(currentSelection), favorites);
    await fetch(`http://localhost:5000/deletefavorite?user_id=1&selection_id=${currentSelection}`);
    dispatch({
        type: SET_FAVORITE,
        payload: newFavorite
    });
} else {
    const newFavorite = append(currentSelection, favorites)
    await fetch(`http://localhost:5000/addfavorite?user_id=1&selection_id=${currentSelection}`);
    dispatch({
        type: SET_FAVORITE,
        payload: newFavorite
    });
}