-1

链接到 TypeError 屏幕截图

该应用程序昨天运行良好。我删除了项目并从 Github.com 克隆。还是不行。这是代码的链接: 链接到 Github 项目

import React from 'react';

function SearchMovies(props) {
  const FavoriteComponent = props.favoriteComponent;
  return (
    <>
      {props.movies.map((movie, index) => (
        <div>
          s<img src={movie.Poster} alt='movie'></img>
          <div onClick={() => props.handleFavoritesClick(movie)}>
            <FavoriteComponent />
          </div>
        </div>
      ))}
    </>
  );
}

export default SearchMovies;
4

1 回答 1

1

在您的SearchMovies组件中尝试使用空传播来避免该错误。

{props.movies?.map((movie, index) => (...

import React from 'react';

function SearchMovies(props) {
  const FavoriteComponent = props.favoriteComponent;
  return (
    <>
      {props.movies?.map((movie, index) => (
        <div>
          s<img src={movie.Poster} alt='movie'></img>
          <div onClick={() => props.handleFavoritesClick(movie)}>
            <FavoriteComponent />
          </div>
        </div>
      ))}
    </>
  );
}

export default SearchMovies;

更正后的工作应用程序:Stackblitz

于 2020-12-29T22:01:55.890 回答