Question

在我的Android应用程序中，我使用的是sqlite数据库但是当使用查询从SQLite读取数据时，它给了我错误。这是我创建表的代码。

package com.example.mydb;
public class Dbadapter {

private static final String DATABASE_NAME = "mydb";
private static final int DATABASE_VERSION = 5;

private static final String DATABASE_TABLE18 = "modalation_medicine_table";
private static final String MOD_ID = "modalation_previousss_id";
private static final String MOD_MED_ID = "modalation_medicine_id";
private static final String MOD_MEDICINE = "modalation_medicine";

private static final String CREATE_DATABASE_TABLE18 = "CREATE TABLE IF NOT EXISTS "
        + DATABASE_TABLE18
        + " ("
        + MOD_MED_ID
        + " INTEGER PRIMARY KEY AUTOINCREMENT, "
        + MOD_MEDICINE
        + " text, "
        + MOD_ID
        + " integer, "
        + " FOREIGN KEY ("
        + MOD_ID
        + ") REFERENCES "
        + DATABASE_TABLE5
        + " ("
        + MODALATION_KEY_ID
        + "));";

private DbHelper ourHelper;
private final Context ourContext;
private SQLiteDatabase ourDatabase;

private static class DbHelper extends SQLiteOpenHelper {

    public DbHelper(Context context) {
        super(context, DATABASE_NAME, null, DATABASE_VERSION);
        // TODO Auto-generated constructor stub
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
          db.execSQL(CREATE_DATABASE_TABLE18);

    ContentValues modalation_medicine = new ContentValues();
        modalation_medicine.put(MOD_MEDICINE, "Natrum Muriaticum");
        modalation_medicine.put(MOD_ID, 1);
        db.insert(DATABASE_TABLE18, null, modalation_medicine);

        modalation_medicine.put(MOD_MEDICINE, "Calcarea Carbonica");
        modalation_medicine.put(MOD_ID, 1);
        db.insert(DATABASE_TABLE18, null, modalation_medicine);

        modalation_medicine.put(MOD_MEDICINE, "Phosphorus");
        modalation_medicine.put(MOD_ID, 1);
        db.insert(DATABASE_TABLE18, null, modalation_medicine);
}

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        // TODO Auto-generated method stub
        db.execSQL("DROP TABLE IF EXIST " + DATABASE_TABLE18);

    }

}

public Dbadapter(Context c) {
    ourContext = c;
}

public Dbadapter open() throws SQLException{
    ourHelper = new DbHelper(ourContext);
    ourDatabase = ourHelper.getWritableDatabase();
    return this;
}

public void close() {
    ourHelper.close();
}

现在，当我使用此方法从我的sqLite数据库中读取数据时，它会给出一个错误“无法重新打开已经关闭的对象”。

public boolean mod_app_symps(String l, String modfinalMed) {
        String[] modalation = new String[] { MOD_ID, MOD_MEDICINE };
        Cursor c24;
        boolean iscontain = false;
        c24 = ourDatabase.query(DATABASE_TABLE18, modalation, "modalation_previousss_id = ? AND modalation_medicine = ?",new String[] { l , modfinalMed } , null, null, null);
        if(c24.getCount() > 0) {
            iscontain = true;
        }
        return iscontain;
    }

我试图通过谷歌解决问题，但没有任何帮助！ P.S：我有许多其他方法正在做同样的工作，即从SQLite数据库读取数据，它们都工作正常。该错误仅在此方法上出现。

编辑：此方法正常

public String[] getmodMed(long l) {
        String[] modalation_med = new String[] { MOD_ID, MOD_MEDICINE};
        Cursor c19 = ourDatabase.query(DATABASE_TABLE18, modalation_med, MOD_ID + "='" + l + "'" , null, null, null, null);
        String mod_medicine[]= new String[c19.getCount()];
        int imodmedicine = c19.getColumnIndex(MOD_MEDICINE);
        if(c19 != null){
            c19.moveToFirst();
            while(!c19.isAfterLast()) {
                for(int i =0; i< c19.getCount(); i++) { 
                mod_medicine[i]= c19.getString(imodmedicine);
                c19.moveToNext();
                } }
        }
        return mod_medicine;
    }

Answer 1

这是因为这行代码：

public void close() {
ourHelper.close();
}

您关闭数据库，然后想再次访问它。

Answer 2

我们需要查看更多代码，特别是＆＃34; ourDatabase＆＃34;。听起来它在你运行该方法之前就已经关闭了。尝试在查询之前添加SQLiteDatabase db = ourHelper.getWriteableDatabase，然后针对＆＃34; db＆＃34;

运行查询

public boolean mod_app_symps(String l, String modfinalMed) {
    SQLiteDatabase db = outHelper.getWriteableDatabase();
    String[] modalation = new String[] { MOD_ID, MOD_MEDICINE };
    Cursor c24;
    boolean iscontain = false;
    c24 = db.query(DATABASE_TABLE18, modalation, "modalation_previousss_id = ? AND modalation_medicine = ?",new String[] { l , modfinalMed } , null, null, null);
    if(c24.getCount() > 0) {
        iscontain = true;
    }         
    return iscontain;
}

当从sqlite android读取数据时，“无法重新打开已经关闭的对象”

2 个答案: