Java 将数据库详细信息转换为 JSON 对象
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Converting the Database details to JSON object
提问by Amit Sharad
I have a table with has the columns namely
我有一张表，其中有列
recordID, recordName , titleFeild, titleIDMap, titleId, titleStartDate, titleEndDate, languageId
recordID、recordName、titleFeild、titleIDMap、titleId、titleStartDate、titleEndDate、languageId
Now I have convert the data from above columns to the JSON object data which looks like below
现在我已将数据从上面的列转换为 JSON 对象数据，如下所示
{
"recordId" :10,
"recordName" : "RECORDS",
"records" : [ {
"titleField" : 1,
"titleIDMap" : null,
"titleId" : 500,
"titleStartDate" : "2013-12-22T00:00:00.000+0000",
"titleEndDate" : "2013-12-03T00:00:00.000+0000",
"languageId" : 20
}]
}
Please note that recordsis an array of columns( titleFeild,titleIDMap,titleId,titleStartDate,titleEndDate,languageId)
请注意，记录是一个列数组（titleFeild、titleIDMap、titleId、titleStartDate、titleEndDate、languageId）
The code so far I have developed is
到目前为止我开发的代码是
List<Object[]> objList = dao.getStatus();
Integer result = null;
JSONObject jsonData = new JSONObject();
JSONArray jsonDataArray = new JSONArray();
if(objList!=null && objList.size()>10000)
{
for (Object[] nameObj : objList) {
jsonData.put("", nameObj.get(arg0) );
}
}
How do I construct the JSON Object from the columns data ?
如何从列数据构造 JSON 对象？
采纳答案by PopoFibo
You can easily achieve this with google-gsonlibrary. In simple terms you would have to create a couple of Pojos(with reference to another containin a list of references).
您可以使用google-gson库轻松实现这一点。简单来说，您必须创建几个Pojo（参考包含在参考列表中的另一个）。
Consider RecordIDand RecordNameas Meta Data.
考虑RecordID和RecordName作为元数据。
Create a pojorepresenting this information:
创建一个pojo代表此信息的：
public class DbMetaPojo {
private int recordID;
private String recordName;
private List<Record> records;
public List<Record> getRecords() {
return records;
}
public void setRecords(List<Record> records) {
this.records = records;
}
public String getRecordName() {
return recordName;
}
public void setRecordName(String recordName) {
this.recordName = recordName;
}
public int getRecordID() {
return recordID;
}
public void setRecordID(int recordID) {
this.recordID = recordID;
}
}
Create another pojowith the actual Recordfields:
pojo使用实际Record字段创建另一个：
public class Record {
public int getTitleFeild() {
return titleFeild;
}
public void setTitleFeild(int i) {
this.titleFeild = i;
}
public String getTitleIDMap() {
return titleIDMap;
}
public void setTitleIDMap(String titleIDMap) {
this.titleIDMap = titleIDMap;
}
public int getTitleId() {
return titleId;
}
public void setTitleId(int titleId) {
this.titleId = titleId;
}
public String getTitleStartDate() {
return titleStartDate;
}
public void setTitleStartDate(String titleStartDate) {
this.titleStartDate = titleStartDate;
}
public String getTitleEndDate() {
return titleEndDate;
}
public void setTitleEndDate(String titleEndDate) {
this.titleEndDate = titleEndDate;
}
public int getLanguageId() {
return languageId;
}
public void setLanguageId(int languageId) {
this.languageId = languageId;
}
private int titleFeild;
private String titleIDMap;
private int titleId;
private String titleStartDate;
private String titleEndDate;
private int languageId;
}
Now just a method to populate your POJOswith the relevant data (replace the hardcoding logic with your data retrieve):
现在只是一种POJOs用相关数据填充您的方法（用您的数据检索替换硬编码逻辑）：
public static void main(String... main) {
DbMetaPojo obj = new DbMetaPojo();
obj.setRecordID(10);
obj.setRecordName("RECORDS");
Record record = new Record();
record.setLanguageId(20);
record.setTitleEndDate("2013-12-22T00:00:00.000+0000");
record.setTitleFeild(1);
record.setTitleId(500);
record.setTitleIDMap("SOME NULL");
record.setTitleStartDate("2013-12-22T00:00:00.000+0000");
List<Record> list = new ArrayList<Record>();
list.add(record);
obj.setRecords(list);
Gson gson = new Gson();
String json = gson.toJson(obj);
System.out.println(json);
}
Output is your formed JSON:
输出是您形成的 JSON：
{
"recordID": 10,
"recordName": "RECORDS",
"records": [
{
"titleFeild": 1,
"titleIDMap": "SOME NULL",
"titleId": 500,
"titleStartDate": "2013-12-22T00:00:00.000+0000",
"titleEndDate": "2013-12-22T00:00:00.000+0000",
"languageId": 20
}
]
}
EDIT:
编辑：
To align to your code, you might want to do something like:
要与您的代码保持一致，您可能需要执行以下操作：
List<Object> objList = dao.getStatus();
List<DbMetaPojo> metaList = new ArrayList<DbMetaPojo> ();
if (objList != null && objList.size() > 10000) {
for (Object nameObj : objList) {
DbMetaPojo meta = new DbMetaPojo();
meta.setRecordID(nameObj[0]);
meta.setRecordName(nameObj[0]);
...
...
...
metaList.add(meta);
}
}
回答by Vinayak Pingale
- First of all what you have to do is retrieve the data from the columns of the table using your
DAOand calling a Function fromDAOIMPLwhich in turn will return the list of data(POJOprobably).
- 首先，您需要做的是使用您的
DAO并调用一个函数从表的列中检索数据，该函数DAOIMPL又将返回数据列表（POJO可能）。
Create a map like this which will contain your key value pair for example recordid and value, recordname and value
创建一个这样的映射，其中将包含您的键值对，例如记录 ID 和值、记录名称和值
Map<String,Object> objMap = new HashMap<String,Object>();
objMap.put("recordId", Record.getId());
objMap.put("recordName",Record.getName());
// Now here is the deal create another hashmap here whose key will be records "the key for your second array"
//Put the values in this second hashmap as instructed above and put it as a key value pair.
........
.......
.......
JSONObject JsonObject = JSONObject.fromObject(objMap);//This will create JSON object out of your hashmap.
objJSONList.add(JsonObject);
}
StringBuffer jsonBuffer = new StringBuffer();
jsonBuffer.append("{\"data\": {");
jsonBuffer.append(objJSONList.tostring());
jsonBuffer.append("}");
//jsonBuffer.append(",\"total\":"+ objJSONList.size());// TOTAL Optional
//jsonBuffer.append(",\"success\":true}");//SUCCESS message if using in callback Optional
回答by Tugrul
Create an object which has your attribues. (recordID, recordName , titleFeild, titleIDMap, titleId, titleStartDate, titleEndDate, languageId)
创建一个具有您的属性的对象。（记录ID、记录名称、titleFeild、titleIDMap、titleId、titleStartDate、titleEndDate、languageId）
Get data from dao and convert it to json. It will looks like what you want.
从 dao 获取数据并将其转换为 json。它看起来像你想要的。
Gson gson = new Gson();
// convert java object to JSON format,
// and returned as JSON formatted string
String json = gson.toJson(obj);
回答by user3192244
I think your dao.getStatus() should return a List with Map keys and values. Your key would be column name and value would be content.
我认为你的 dao.getStatus() 应该返回一个带有 Map 键和值的列表。您的键是列名，值是内容。
List<Map<String,Object>> objList = dao.getStatus();
if(objList!=null && objList.size()>10000){
for(Map<String,Object> row : objList) {
Iterator<String> keyList = row.keySet().iterator();
while(keyList.hasNext()){
String key = keyList.next();
jsonData.put(key, row.get(key));
}
}
}
For the records array you need to build it while iterating table columns. Combining above code with building records array would be something like this..
对于记录数组，您需要在迭代表列时构建它。将上面的代码与构建记录数组相结合将是这样的..
String[] group = {"titleField","titleIDMap","titleId","titleStartDate","titleEndDate","languageId"};
List<String> recordGroup = Arrays.asList(group);
Map<Object, JSONArray> records = new HashMap<Object,JSONArray>();
List<Map<String,Object>> objList = dao.getStatus();
JSONObject jsonData = new JSONObject();
if(objList!=null && objList.size()>10000){
for(Map<String,Object> row : objList) {
int columnCount = 0;
Iterator<String> keyList = row.keySet().iterator();
while(keyList.hasNext()){
String key = keyList.next();
if(recordGroup.contains(key)){
Object recordId = row.get("recordId");
JSONArray recordArray = new JSONArray();
if(records.containsKey(recordId)){
recordArray = records.get(recordId);
JSONObject jsonObj = null;
if(columnCount >= recordGroup.size()){
jsonObj = new JSONObject();
recordarray.add(jsonObj);
columnCount = 0;
}
else {
jsonObj = (JSONObject) recordArray.get(recordArray.size()-1);
}
jsonObj.put(key, row.get(key));
columnCount++;
}
else {
JSONObject jsonObj = new JSONObject();
jsonObj.put(key, row.get(key));
recordArray.add(jsonObj);
records.put(recordId, recordArray);
}
jsonData.put("records", records.get(recordId));
}
else {
jsonData.put(key, row.get(key));
}
}
}
}
