光標getString（）或getInt（）不返回給定索引的值

Question

我已經爲Android應用程序創建了SQLite數據庫和表。可以插入值並且返回長ID而不會出現問題。當返回的ID用於檢索其他字段時，getColumnIndex（COLUMN_NAME）返回正確的索引，但是何時使用getSting和getInt獲取字段的值時，應用程序會崩潰。 Android工作室不會給出任何錯誤。請幫忙。

數據庫處理類：

public class SqliteDBManager extends SQLiteOpenHelper{ 
private static final int DB_VERSION = 1; 
private static final String DB_NAME = "baby_diary.db" 

private static final String TBL_BABIES = "babies"; 
private static final String BABIES_BABY_ID = "baby_id"; 
private static final String BABIES_NAME = "name"; 
private static final String BABIES_DOB = "dob"; 
private static final String BABIES_GENDER = "gender"; 
private static final String BABIES_WEIGHT = "weight"; 
private static final String BABIES_HEIGHT = "height"; 
private static final String BABIES_THEME = "theme"; 

public SqliteDBManager(Context context, String name, SQLiteDatabase.CursorFactory factory, int version, DatabaseErrorHandler errorHandler) { 
     super(context, DB_NAME, factory, DB_VERSION, errorHandler); 
    } 

public void onCreate(SQLiteDatabase db) { 

     String CREATE_TBL_BABIES = "CREATE TABLE " + TBL_BABIES + "(" + 
       BABIES_BABY_ID + " INTEGER PRIMARY KEY AUTOINCREMENT, " + 
       BABIES_NAME + " TEXT, " + 
       BABIES_DOB + " TEXT, " + 
       BABIES_GENDER + " TEXT, " + 
       BABIES_WEIGHT + " REAL, " + 
       BABIES_HEIGHT + " REAL, " + 
       BABIES_THEME + " TEXT" + ");"; 

       db.execSQL(CREATE_TBL_BABIES); 
    } 
    public Babies getBabyById(Long id){ 
     SQLiteDatabase db = getWritableDatabase(); 
     String query = "SELECT * FROM " + TBL_BABIES + " WHERE " + BABIES_BABY_ID 
       + " = \"" + id + "\";"; 
     Cursor cursor = db.query(TBL_BABIES, 
       null, 
       BABIES_BABY_ID + " = " + id, 
       null, null, null, null); 

     if (cursor != null && cursor.getCount() > 0) { 
       //for testing getInt 
       int id = cursor.getInt(cursor.getColumnIndex(BABIES_BABY_ID)); 
       String name = cursor.getString(cursor.getColumnIndex(BABIES_NAME)); 
       cursor.close(); 
      } 

Answer 1

你必須從遊標查詢的第一行之前使用cursor.moveToFirst（）。將您的if語句更改爲：

if (cursor != null && cursor.moveToFirst()) { 

而且，即使沒有結果，也必須關閉遊標。我建議你這樣寫：

try { 
    if (cursor != null && cursor.moveToFirst) { 
      //for testing getInt 
      int id = cursor.getInt(cursor.getColumnIndex(BABIES_BABY_ID)); 
      String name = cursor.getString(cursor.getColumnIndex(BABIES_NAME)); 
     } 
} 
finally { 
    if(cursor != null) { 
     cursor.close(); 
    } 
} 

這樣，無論發生什麼異常，你的光標都會被正確關閉。

Answer 2

我認爲你的光標沒有數據。

Cursor cursor = db.query(TBL_BABIES, 
      null, //This must not null. In here your columns in string array 
      BABIES_BABY_ID + " = " + id, 
      null, null, null, null); 

例如：

Cursor cursor = db.query(TBL_BABIES, 
      new String[]{BABIES_BABY_ID, BABIES_NAME}, //What column you need 
      BABIES_BABY_ID + " = " + id, 
      null, null, null, null); 

在somecase你需要內部if條件添加此命令

cursor.moveToFirst();