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从 Remix.run 中的服务器获取数据

[英]Fetching data from server in Remix.run

I was exploring Remix.run and making a sample app.我正在探索Remix.run并制作一个示例应用程序。 I came across an issue that has been bothering me for some time now.我遇到了一个困扰我一段时间的问题。 Correct me if I am wrong: action() is for handling Form submission, and loader() is for fetching data initially.如果我错了,请纠正我: action()用于处理表单提交,而loader()用于最初获取数据。 For my application, I used mongoose to connect MongoDB, defined models, and defined querying function in a query.server.ts file.对于我的应用程序,我使用 mongoose 连接 MongoDB,定义模型,并在query.server.ts文件中定义查询 function。 I want to fetch data from the database through a function defined in the query.server.ts file when an image is clicked on the UI.当在 UI 上单击图像时,我想通过 q query.server.ts文件中定义的 function 从数据库中获取数据。 How can I do that without using forms?如果不使用 forms,我该怎么做? I cannot pre-fetch the data without knowing what image was clicked by the user.我无法在不知道用户点击了什么图像的情况下预取数据。

You can create a resource route.您可以创建资源路由。 These are like regular routes, but don't export a default component (no UI).这些类似于常规路由,但不导出默认组件(无 UI)。

You can use the useFetcher hook and call fetcher.load() to call your resource route.您可以使用useFetcher挂钩并调用fetcher.load()来调用您的资源路由。 The data is in fetcher.data .数据在fetcher.data中。

// routes/query-data.ts
export const loader: LoaderFunction = async ({request}) => {
  const url = new URL(request.url)
  const img = url.searchParams.get('img')
  const data = await getData(img)
  return json(data)
}

// routes/route.tsx
export default function Route() {
  const fetcher = useFetcher()

  const handleImgClick = (e) => {
    const img = e.target
    fetcher.load(`/query-data?img=${img.attr('src')}`)
  }

  return (
    <div>
      <img onClick={handleImageClick} src="/images/file.jpg" />
      <pre>{ JSON.stringify(fetcher.data, null, 2) }</pre>
    </div>
  )
}

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