Question

如何在JSONArray内解析JSONObject？以下是我从服务器获取的JSON响应。

{
"searchdata": {
    "titles": [
        "<b>Laptop</b> - Wikipedia, the free encyclopedia",
        "<b>laptop</b> - definition of <b>laptop</b> by the Free Online Dictionary ..."
    ],
    "desc": [
        "A <b>laptop</b> computer is a personal computer for mobile use. A <b>laptop</b> has most of the same components as a desktop computer, including a display, a keyboard, a ...",
        "lap·top (l p t p) n. A portable computer small enough to use on one&apos;s lap. <b>laptop</b> [ˈlæpˌtɒp], <b>laptop</b> computer. n (Electronics &amp; Computer Science / Computer ..."
    ],
    "links": [
        "http://en.wikipedia.org/wiki/Laptop",
        "http://www.thefreedictionary.com/laptop"
    ],
    "nextpage": ""
}
}

我能够获得JSONObject但是如何逐个获取JSONArray，以便我可以在listview中修复它们。

我想在listview的一行中显示每个数组的值，依此类推......

任何帮助将不胜感激。

Answer 1

非常容易..

你需要修改这样的代码：

//jsonString is your whole JSONString which you have shown above

JSONObject jObj = new JSONObject(jsonString);
JSONObject jSearchData = jObj.getJSONObject("searchdata");
JSONArray jTitles = jSearchData.getJSONArray("titles");
JSONArray jDesc= jSearchData.getJSONArray("desc");
JSONArray jLinks= jSearchData.getJSONArray("links");
String nextPage = jSearchData.getString("nextpage");
//and so on and so forth

用于获取数组项并将其显示在列表视图中：

//you can iterate over each item and add it to an ArrayList like this:

//i am showing you a single one,follow the same process for other arrays:

ArrayList titlesArray = new ArrayList();

for (int i = 0; i < jTitles.length(); i++) {
String item = jTitles.getString(i);
titlesArray.add(item);

}

接下来，你让这个arraylist成为这样的listview的源：

 // Get a handle to the list view
    ListView lv = (ListView) findViewById(R.id.ListView01);
 lv.setAdapter(new ArrayAdapter<string>((Your activity class).this,
            android.R.layout.simple_list_item_1, titlesArray));

Answer 2

考虑将您的顶级JSON解析为JSONObject，随后您可以通过方法getJSONObject(name)和{{请求从中接收任何子级别对象和/或数组1}}。您感兴趣的数组在JSON层次结构中有两个级别，所以您需要这样的内容：

getJSONArray(name)

您可以这样迭代任何数组（以String json = ...; JSONObject rootObj = new JSONObject(json); JSONObject searchObj = rootObj.getJSONObject("searchdata"); JSONArray titlesObj = searchObj.getJSONArray("titles"); JSONArray descsObj = searchObj.getJSONArray("desc"); JSONArray linksObj = searchObj.getJSONArray("links");为例）：

titles

Answer 3

这将有所帮助：

JSONArray  titles = new jSONObject(jsonstring).getJSONObject("searchdata").getJSONArray("titles");

JSONArray  desc  = new jSONObject(jsonstring).getJSONObject("searchdata").getJSONArray("desc");

JSONArray  links = new jSONObject(jsonstring).getJSONObject("searchdata").getJSONArray("links");

解析JSONObject和JSONArray并在listview中修复它们

3 个答案: