将 JSONObject 传递给另一个活动 [英] Passing JSONObject into another activity

问题描述

我正在使用一个返回 JSON 数据(新 DVD 标题)的外部 API.我能够解析出 JSON 并将每个 dvd 标题和其他 dvd 信息列出到 ListView 中.我还能够为原始数据(标题字符串)设置一个 onListItemClick 方法.我最终为 onListItemClick 方法编写了类似的内容:

I'm hitting an external API that's returning JSON data (new dvd titles). I'm able to parse out the JSON and list each dvd title and other dvd information into a ListView just fine. I was also able to setup an onListItemClick method just fine for primitive data (title string). I ended up writing something like so for the onListItemClick method:

请注意，productArray 是一个类 var，它由另一个包含 JSONObjects 数组的方法设置.

Just to note, the productArray is a class var that's being set by another method that holds an array of JSONObjects.

protected void onListItemClick(ListView l, View v, int position, long id) {
    super.onListItemClick(l, v, position, id);

    Intent i = new Intent(DvdListingActivity.this, MovieProductActivity.class);
    try {
        JSONObject jsonObj = productArray.getJSONObject(position);
        i.putExtra("mTitle", jsonObj.getJSONObject("Title").opt("val").toString());
        i.putExtra("mRelDate", jsonObj.getJSONObject("RelDate").opt("val").toString());
        i.putExtra("mDesc", jsonObj.getJSONObject("Desc").opt("val").toString());
        i.putExtra("mRating", jsonObj.getJSONObject("MPAA").getJSONObject("Rating").opt("val").toString());
        i.putExtra("mActors", jsonObj.getJSONObject("Actors").opt("val").toString());
        i.putExtra("mImage", jsonObj.getJSONObject("Image").opt("val").toString());
        startActivity(i);

    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }       

}

以上代码都有效，但我认为 GOTTA 是我将数据传递给另一个 Activity 的更好方法.我在想我将能够传递一个包含 DVD 电影的所有数据的 JSONObject，而不是单独设置每个数据点.

The above code all works, but I'm thinking there's GOTTA be a better way for me to pass in data to another Activity. I was thinking that I would be able to pass a JSONObject that contains all the data for a dvd movie instead of setting each data point individually.

我尝试了一个半星期来弄清楚如何使用 Parcelable.我尝试实例化一个新的 JSONObject jsonObj，它实现了 Parcelable，但没有运气.我的 LogCat 中不断出现错误，指出该对象不可打包.

I tried for a week and a half to figure out how to use Parcelable. I tried instantiating a new JSONObject jsonObj that implements Parcelable with no luck. I kept getting an error in my LogCat that said that the object was un-parcelable.

我尝试阅读 Android 开发者网站和其他博客，但我无法将他们的示例应用到我需要做的事情上.

I've tried reading the Android developer site and other blogs, but I couldn't apply their examples to what I needed to do.

任何帮助将不胜感激

推荐答案

你可以将一部电影的所有信息封装到一个Movie对象中，该对象实现了Parcelable.

You can just encapsulate all of the information about a movie into a Movie object, which implements Parcelable.

代码看起来与上面类似，但您可以只传递一个电影，而不是传递 6 个不同的附加内容.

The code will look similar to above, but instead of passing 6 different extras you can just pass one extra that is the movie.

Movie movie = new Movie();
movie.setTitle(jsonObj.getJSONObject("Title").opt("val").toString());
movie.setRelDat(jsonObj.getJSONObject("RelDate").opt("val").toString());
.
.
.
i.putExtra("movie", movie);

有关实现 Parcelable 对象的信息，请参阅 Parcelable 文档.您基本上只需写出 'writeToParcel' 中的每个字符串，然后按正确顺序读入 'readFromParcel' 中的每个字符串.

For information on implementing a Parcelable object, see Parcelable docs. You basically just write out each string in 'writeToParcel', and read in each string in 'readFromParcel' in the correct order.

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