Question

我有一个来自服务器的JSON字符串，它看起来像这样：

{
    "categories": {
        "0": {
            "term_id": "247",
            "name": "Content Curation"
        },
        "1": {
            "term_id": "50",
            "name": "Content Marketing"
        },
        "2": {
            "term_id": "2",
            "name": "Curation"
        },
        "3": {
            "term_id": "246",
            "name": "Inbound Marketing"
        },
        "4": {
            "term_id": "47",
            "name": "Marketing"
        },
        "5": {
            "term_id": "4",
            "name": "News Curation"
        },
        "6": {
            "term_id": "36",
            "name": "SEO"
        },
        "8": {
            "term_id": "248",
            "name": "Wordpress Content Curation"
        }
    }
}

我的任务是获取＆＃34; term_id＆＃34;的值。和＆＃34;名称＆＃34;领域。我曾经习惯于获得＆＃34;类别＆＃34;使用此代码从当前JSONObject对象：

JSONObject jObject = new JSONObject(responceData);
JSONObject categoryObject = jObject.getJSONObject("categories");
JSONArray  jarray = new JSONArray("["+categoryObject.toString().substring(1,categoryObject.toString().length() - 1) + "]");

for(int i=0;i<jarray.length();i++)
{
     JSONObject jobj = jarray.getJSONObject(i);
     String term_id=jobj.getString("term_id");
     String name=jobj.getString("name");
}

并且categoryObject看起来像这样：

{
    "0": {
        "term_id": "247",
        "name": "Content Curation"
    },
    "1": {
        "term_id": "50",
        "name": "Content Marketing"
    },
    "2": {
        "term_id": "2",
        "name": "Curation"
    },
    "3": {
        "term_id": "246",
        "name": "Inbound Marketing"
    },
    "4": {
        "term_id": "47",
        "name": "Marketing"
    },
    "5": {
        "term_id": "4",
        "name": "News Curation"
    },
    "6": {
        "term_id": "36",
        "name": "SEO"
    },
    "8": {
        "term_id": "248",
        "name": "Wordpress Content Curation"
    }
}

但在那之后，我不知道如何获得田地。有没有办法从JSONObject获取所有JSONObject子项？

如果您有源代码或可以举个例子，请与我分享。

Answer 1

在这里你可以检索你所有的json数据，要求一个特定的密钥和innerKey来得到你想要的东西，干杯

    try
    {   
        String jsonString="";//your json string here
        JSONObject jObject= new JSONObject(jsonString).getJSONObject("categories");
        Iterator<String> keys = jObject.keys();
        while( keys.hasNext() )
        {
            String key = keys.next();
            Log.v("**********", "**********");
            Log.v("category key", key);
            JSONObject innerJObject = jObject.getJSONObject(key);
            Iterator<String> innerKeys = innerJObject.keys();
            while( innerKeys.hasNext() )
            {
                String innerKkey = keys.next();
                String value = innerJObject.getString(innerKkey);
                Log.v("key = "+key, "value = "+value);
            }
        }
    }
    catch (JSONException e)
    {   e.printStackTrace();    }

Answer 2

检查this answer，这样你不需要知道密钥，并且可以循环访问内部对象，但是类别会更好地作为JSONArray而不是Object，这样你就可以只是正常循环

for(int i = 0; i < array.length(); i++) {
    JSONObject obj = myArray.get(i);
    ....
}

Answer 3

非常感谢Ahmad Dwaik和Tom Hart的回答。这是解决方案的代码。

try
        {   

            JSONObject jObject= new JSONObject(responseData).getJSONObject("categories");
            Iterator<String> keys = jObject.keys();
            while( keys.hasNext() )
            {
                String key = keys.next();
                Log.v("**********", "**********");
                Log.v("category key", key);
                JSONObject innerJObject = jObject.getJSONObject(key);

                String name = innerJObject.getString("name");
                String term_id = innerJObject.getString("term_id");

                Log.v("name = "+name, "term_id = "+term_id);

            }
        }
        catch (JSONException e){
           e.printStackTrace();    
        }

Answer 4

使用keys（）迭代器迭代所有属性，并为每个属性调用get（）。

Iterator<String> iter = json.keys();
while (iter.hasNext()) {
    String key = iter.next();
    try {
        Object value = json.get(key);
    } catch (JSONException e) {
        // Something went wrong!
    }
}

请参阅：https://stackoverflow.com/a/13573965/2480911

Answer 5

试试这个

JSONObject jObject = new JSONObject(responceData);
JSONObject categoryObject = jObject.getJSONObject("categories");
JSONObject obj0 = categoryObject.getJSONObject("0");
String termId0 obj0.getString("term_id");
String name0 obj0.getString("name");
JSONObject obj1 = categoryObject.getJSONObject("1");
String termId1 obj1.getString("term_id");
String name1 obj1.getString("name");

但我同意wqrahd，类别应该是一个数组

Answer 6

Ahmad Dwaik＆＃39; Warlock＆＃39;提供了一个非常好的答案，但他的代码当前代码有一个错误 -

Dim i As Long
Dim market As Variant, arrbooks() As Workbook

Application.ScreenUpdating = False
Application.DisplayAlerts = False
Sheet1.AutoFilterMode = False
market = Array(...
 ReDim arrbooks(0 To UBound(market))

'create workbooks
For i = 0 To UBound(market)
    Set arrbooks(i) = Workbooks.Add
Next

'retrieve data by autofilter

With Sheet4

    For i = 0 To UBound(market)

        .Range("H:H").AutoFilter field:=1, Criteria1:=market(i)

        .Range("H1").CurrentRegion.SpecialCells(xlCellTypeVisible).Copy

       Workbooks(arrbooks(i).Name).Sheets(1).Range("A1").PasteSpecial Paste:=xlPasteAll


**workbooks(arrbooks(i).name).sheets(1).application.cutcopymode = false**

        Next
End With

'save workbooks  

For i = 0 To UBound(market)

**workbooks(arrbooks(i).name).sheets(1).application.cutcopymode = false**

Workbooks(arrbooks(i).Name).SaveAs "insert save path"
Workbooks(arrbooks(i).Name).Close
Next

'clean up
Application.ScreenUpdating = False
Sheet4.ShowAllData
Sheet4.AutoFilterMode = False
Application.DisplayAlerts = True

Sheet4.Activate
Sheet4.Range("A1").Select

Answer 7

尝试->

String response = data;
JSONObject categories= new JSONObject(data).getJSONObject("categories").getJSONObject('0').getJSONObject('term_id');

打印// 247

在JSONObject中解析多个JSONObject的Android JSON

7 个答案: