Question

用于json解析和迭代的API：

org.json.simple
org.json

我正在尝试从ArchiveBatchChangeRequest下的FieldName中获取两个值 - ＆gt; DocumentSets - ＆gt; QuerySpecification并将其作为ArrayList返回。

elementNode = “ArchiveBatchChangeRequest; DocumentSets; QuerySpecification”
objectValue =“FieldName”
batchFile = JSON输入文件

因此返回值应包含CustomerName和DateCreated，因为QuerySpecification是一个大小为2的数组

我似乎无法让这个工作。我担心我的起点非常糟糕，或者可以做得更容易。

JSON输入文件：

{
    "ArchiveBatchChangeRequest": [{
        "BatchRunSpecification": [{
            "BatchOperation": "EXTRACT"
        }]
    },
    {
        "OutputSpecification": [{
                "ReportFieldNames": ["ReportField1"]
        }]  
    },
    {
        "DocumentSets": [{
            "DocumentSetNo": "1",
            "QuerySpecification": [{
                "FieldName": "CustomerName",
                "SimpleQuery": [{
                    "Operator": "EQUAL",
                    "Values": ["Customer1"]
                }]
            },
            {
                "FieldName": "DateCreated",
                "SimpleQuery": [{
                    "Operator": "EQUAL",
                    "Values": ["19-12-2015"]
                }]
            }]
        }]
    }]
}

代码：

        public ArrayList<String> getElementValue(String elementNode, String objectValue, File batchFile) throws IOException, ParseException{
            String [] elementArray = elementNode.split(";");
            JSONObject outerObject = null;
            ArrayList<String> values = new ArrayList<String>();
            try {
                JSONParser parser = new JSONParser();
                    Object object = parser.parse(new FileReader(batchFile));
                outerObject = new JSONObject(object.toString());
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (FileNotFoundException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            JSONArray getArray = null;

            for (int i = 0; i < elementArray.length; i++) {
                getArray = outerObject.getJSONArray(elementArray[i]);

                for (int j = 0; j < getArray.length(); j++) {
                    for (int j2 = 0; j2 < elementArray.length; j2++) {
                        if(getArray.getJSONObject(j).names().get(0).equals(elementArray[j2])){
                            outerObject = getArray.getJSONObject(j);
                        }
                    }
                }

                if(i == elementArray.length - 1){
                    outerObject = getArray.getJSONObject(0);
                }
            }

            try{
                getArray = null;
                getArray = outerObject.getJSONArray(objectValue);
            }catch(JSONException je){ //Object is not an JSONArray
                je.printStackTrace();
            }finally{
                if(getArray != null){
                    for (int i = 0; i < getArray.length(); i++) {
                        values.add(getArray.getString(i).replace("\"", ""));
                    }
                }else 
                    values.add(outerObject.getString(objectValue));
            }


            return values;
        }

Answer 1

这是您的任务和使用方式..

public void parse(String data) throws JSONException{  
    org.json.JSONObject jsonObject=new org.json.JSONObject(data);  
    JSONArray array=jsonObject.getJSONArray("ArchiveBatchChangeRequest");  
    for(int i=0;i<array.length();i++){  
        org.json.JSONObject jsonObject2=array.getJSONObject(i);  
        if(i==0){  
        JSONArray     array1=jsonObject2.getJSONArray("BatchRunSpecification");
        org.json.JSONObject jsonObject3=array1.getJSONObject(0);  
          System.out.println("BatchOperation="+jsonObject3.get("BatchOperation"));  
        }else if(i==1){
            JSONArray array1=jsonObject2.getJSONArray("OutputSpecification");
            org.json.JSONObject jsonObject3=array1.getJSONObject(0);

            JSONArray array2=jsonObject3.getJSONArray("ReportFieldNames");

            System.out.println("ReportFieldNames="+array2.getString(0));
        }
        else if(i==2){

            JSONArray array1=jsonObject2.getJSONArray("DocumentSets");
            org.json.JSONObject jsonObject3=array1.getJSONObject(0);
            System.out.println("DocumentSetNo="+jsonObject3.getString("DocumentSetNo"));
            JSONArray array2=jsonObject3.getJSONArray("QuerySpecification");
            for(int j=0;j< array2.length();j++){

                    org.json.JSONObject jobject=array2.getJSONObject(j);
                    String fieldName=jobject.getString("FieldName");
                    System.out.println("value fo FieldName use as you want"+fieldName);
                    String operator=jobject.getJSONArray("SimpleQuery").getJSONObject(0).getString("Operator");
                    System.out.println("value fo operator use as you want-"+operator);
                    String values=jobject.getJSONArray("SimpleQuery").getJSONObject(0).getJSONArray("Values").getString(0);
                    System.out.println("value fo values use as you want-"+values);

            }


        }


    }

}

我希望它会帮助你......！

在java中获取JSON值

1 个答案: