JAVA/ANDROID 数组转换为LIST

Posted 2023-05-04

我有一个数组, String[][] datas = "aaa","aa1","aa11","bbb","bb1","bb11","ccc","cc1","cc11",
"aaa","aa2","aa22","bbb","bb2","bb22","aaa","aa3","aa33",
"bbb","bb3","bb33";

public class DetailModel
public String pron = "";

public List<MeanPhrase> mpList;

//pron = "aaa", mean="aa1", phrase="aa11"

public class MeanPhrase
public String mean, phrase;


想按照"aaa","bbb","ccc"来分组, 转为相应的 LIST<DetailModel>

参考技术A String[][] datas = "aaa","aa1","aa11","bbb","bb1","bb11","ccc","cc1","cc11",
"aaa","aa2","aa22","bbb","bb2","bb22","aaa","aa3","aa33",
"bbb","bb3","bb33";

Map<String , DetailModel> map = new HashMap<String , DetailModel>();
for (String[] strings : datas)

DetailModel detail = new DetailModel();

detail.setPron(strings[0]);

List<MeanPhrase> li = new ArrayList<MeanPhrase>();
MeanPhrase mean = new MeanPhrase();
mean.setMean(strings[1]);
mean.setPhrase(strings[2]);
li.add(mean);

detail.setMpList(li);

//如果已存在相同的key，如aaa,则在key对应的对象的list中增加元素
if(map.containsKey(detail.getPron()))
map.get(detail.getPron()).getMpList().addAll(li);
else//否则直接新增
map.put(detail.getPron(), detail);


//将map转为list
List<DetailModel> resultList = new ArrayList<DetailModel>(map.values());

for (DetailModel detailModel : resultList)
System.out.println(detailModel);

//输出map中的值以及分组结果记录数
System.out.println("======================================");
Set<String> keySet = map.keySet();
for (String string : keySet)
System.out.println(string+"："+map.get(string).getPron()+"："+map.get(string).getMpList().size());


控制台输出结果：
DetailModel [pron=aaa, mpList=[MeanPhrase [mean=aa1, phrase=aa11], MeanPhrase [mean=aa2, phrase=aa22], MeanPhrase [mean=aa3, phrase=aa33]]]
DetailModel [pron=ccc, mpList=[MeanPhrase [mean=cc1, phrase=cc11]]]
DetailModel [pron=bbb, mpList=[MeanPhrase [mean=bb1, phrase=bb11], MeanPhrase [mean=bb2, phrase=bb22], MeanPhrase [mean=bb3, phrase=bb33]]]
======================================
aaa：aaa：3
ccc：ccc：1
bbb：bbb：3

注意：DetailModel [pron=ccc, mpList=[MeanPhrase [mean=cc1, phrase=cc11]]]输出这样的形式是因为我实体类重写了toString()方法的原因，你那不重写的话输出的应该只是对象地址。本回答被提问者采纳 参考技术B

public class test

public static String[][] datas = "aaa", "aa1", "aa11",

"bbb", "bb1", "bb11", "ccc", "cc1", "cc11",

"aaa", "aa2", "aa22", "bbb", "bb2", "bb22",

"aaa", "aa3", "aa33", "bbb", "bb3", "bb33";

public static List<MeanPhrase> mpList;

// pron = "aaa", mean="aa1", phrase="aa11"

public static class MeanPhrase

public String pron, mean, phrase;

public MeanPhrase(String pron, String mean, String phrase)

this.pron = pron;

this.mean = mean;

this.phrase = phrase;

public static void main(String[] args)

for (int i = 0; i < datas.length; i++)

MeanPhrase meanPhrase = new MeanPhrase(datas[i][0],datas[i][1],datas[i][2]);

mpList.add(meanPhrase);

参考技术C 没什么简便方法 遍历这个数组然后封装到DetailModel类里面，封装好一个就用list.add加一个 参考技术D datas转String
1、下载个Google的gson包，
Gson gson = new Gson();
List<MeanPhrase> list = gson.fromJson(str, new TypeToken<List<MeanPhrase>>().getType());
2、下在json包
使用JSONObject.

java android sqlite将完整的表转换为数组

getAllContacts()
    // Getting All Contacts
 public List<Contact> getAllContacts() {
    List<Contact> contactList = new ArrayList<Contact>();
    // Select All Query
    String selectQuery = "SELECT  * FROM " + TABLE_CONTACTS;
 
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
 
    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Contact contact = new Contact();
            contact.setID(Integer.parseInt(cursor.getString(0)));
            contact.setName(cursor.getString(1));
            contact.setPhoneNumber(cursor.getString(2));
            // Adding contact to list
            contactList.add(contact);
        } while (cursor.moveToNext());
    }
 
    // return contact list
    return contactList;
}