[英]How to retrieve a single column data from a Android sugar orm database
我已經在我的應用程序中成功創建了Sugar ORM
數據庫,我可以更新,刪除並從一行中獲取所有數據,但是我希望將單列數據與另一個數據進行匹配...
我的意思是,我有一個注冊數據庫,其中包含以下字段: username
, password
, first_name
, last_name
, email
字段。
使用正確的用戶名和密碼登錄用戶后,我希望將Textview
中該用戶的First_Name發送給Next Activity
...我該怎么做? 在過去的兩天里,我嘗試了但失敗了,請幫助我。
public static List<String> getResultWithRawQuery(String rawQuery, Context mContext) {
List<String> stringList = new ArrayList<>();
if (mContext != null) {
long startTime = System.currentTimeMillis();
SugarDb sugarDb = new SugarDb(mContext);
SQLiteDatabase database = sugarDb.getDB();
try {
Cursor cursor = database.rawQuery(rawQuery, null);
try {
if (cursor.moveToFirst()) {
do {
stringList.add(cursor.getString(0));
} while (cursor.moveToNext());
}
Timber.d(cursor.getString(0), "hi");
} finally {
try {
cursor.close();
} catch (Exception ignore) {
}
}
} catch (Exception e) {
e.printStackTrace();
}
long endTime = System.currentTimeMillis();
long totalTime = endTime - startTime;
System.out.println("total time query" + totalTime);
}
return stringList;
}
另一個示例在列中返回值列表。 這樣使用: String rawQuery = ("SELECT feed_key FROM team_feed_key WHERE team_id = " + mTeam_id + " ORDER BY feed_key DESC");
您是否嘗試過運行像這樣的原始查詢?
List<Note> notes = Note.findWithQuery(Note.class, "Select * from Note where name = ?", "satya");
您可以將功能永久添加到SugarRecord.java
public static String Scaler(String Query) {
String Result = "";
SugarDb db = getSugarContext().getSugarDb();
SQLiteDatabase sqLiteDatabase = db.getDB();
SQLiteStatement sqLiteStatament = sqLiteDatabase
.compileStatement(Query);
try {
Result = sqLiteStatament.simpleQueryForString();
} catch (Exception e) {
e.printStackTrace();
} finally {
sqLiteStatament.close();
}
return Result;
}
要么
public static String Scaler(String Query) {
String Result = "";
SQLiteDatabase sqLiteDatabase = SugarContext.getSugarContext().getSugarDb().getDB();
SQLiteStatement sqLiteStatament = sqLiteDatabase
.compileStatement(Query);
try {
Result = sqLiteStatament.simpleQueryForString();
} catch (Exception e) {
e.printStackTrace();
} finally {
sqLiteStatament.close();
}
return Result;
}
縮放器(“從注釋中選擇名字,其中名稱='ali'限制1“);
我有同樣的問題。 我希望這可以幫助別人:
String firstName = Select.from(User.class).where("EMAIL = "+ user.getEmail()).first().getFirstName();
嗨,這必須可行,您不能編輯庫,但是可以擴展它們,所以請檢查以下內容:
public class DBUtils extends SugarRecord {
public static <T> List<Object> findByColumn(Context context, String tableName,T ColumnObjectType, String columnName) {
Cursor cursor = new SugarDb(context).getDB().query(tableName, new String[]{columnName}, null, null,
null, null, null, null);
List<Object> objects = new ArrayList<>();
while (cursor.moveToNext()){
if (ColumnObjectType.equals(long.class) || ColumnObjectType.equals(Long.class)) {
objects.add(cursor.getLong(0));
}else if(ColumnObjectType.equals(float.class) || ColumnObjectType.equals(Float.class)){
objects.add(cursor.getFloat(0));
}else if(ColumnObjectType.equals(double.class) || ColumnObjectType.equals(Double.class)){
objects.add(cursor.getDouble(0));
}else if(ColumnObjectType.equals(int.class) || ColumnObjectType.equals(Integer.class)){
objects.add(cursor.getInt(0));
}else if(ColumnObjectType.equals(short.class) || ColumnObjectType.equals(Short.class)){
objects.add(cursor.getShort(0));
}else if(ColumnObjectType.equals(String.class)){
objects.add(cursor.getString(0));
}else{
Log.e("SteveMoretz","Implement other types yourself if you needed!");
}
}
if (objects.isEmpty()) return null;
return objects;
}
}
用法很簡單,使用DBUtils.findByColumn(...); 從現在開始,任何您喜歡的地方都只能使用此類而不是SugarRecord,還可以添加自己的其他功能。
提示:ColumnObjectType作為名稱,就像您發送Integer.class一樣,建議告訴列的類型
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