SQlite返回0

让我首先解释如何从SQLite插入和返回数据.

首先,我创建这样的表:

private static final String CREATE_TABLE_STUDENT_LESSON = " create table STUDENTSPECIFICLESSON ( _id TEXT , _viewID INTEGER PRIMARY KEY AUTOINCREMENT , _LESSON_TITLE TEXT , _LESSON_DATE TEXT , _LESSON_PROBLEM TEXT );";

public void onCreate(SQLiteDatabase db) {
    db.execSQL(CREATE_TABLE_STUDENT_LESSON);
}

然后我在SQLite中插入这样的表:

public void insertLesson(String _id, String lessonTitle, String lessonDate, String lessonProblem) {
    ContentValues cOntentValue= new ContentValues();
    contentValue.put(SQLiteHelper._ID, _id);
    contentValue.put(SQLiteHelper.LESSON_TITLE, lessonTitle);
    contentValue.put(SQLiteHelper.LESSON_DATE, lessonDate);
    contentValue.put(SQLiteHelper.LESSON_PROBLEM, lessonProblem);
    this.getWritableDatabase().insert(SQLiteHelper.TABLE_NAME_STUDENTSPECIFICLESSON, null, contentValue);
}

我从Activity中插入它,如下所示:

sqLiteHelper.insertLesson(id,etLessonTitle.getText().toString(),currentDateandTime,etLessonProbStu.getText().toString());

您会注意到我创建了_viewID INTEGER PRIMARY KEY AUTOINCREMENT我正在使用它来获取我的适配器中的特定ViewHolder.

我_viewID通过这样做得到这个:

public String getLessonViewHolderID(String _path){
    String id = null;

    Cursor cursor = getReadableDatabase().rawQuery("Select "+SQLiteHelper._viewID+" from "+SQLiteHelper.TABLE_NAME_STUDENTSPECIFICLESSON+" Where "
            +SQLiteHelper.LESSON_TITLE +"='"+_path+"'",null);

    if (cursor != null) {
        cursor.moveToFirst();
    }
    if (cursor == null) {
    } else if (cursor.moveToFirst()) {
        do {
            id = String.valueOf(cursor.getInt(cursor.getColumnIndex(SQLiteHelper._viewID)));

        } while (cursor.moveToNext());
        cursor.close();
    } else {

    }
    return id;
}

这完美_viewId地工作并按预期正确返回.

然后我尝试_LESSON_PROBLEM通过这样做来从SQLite:

public String getLessonProblem(String _id){


    String id = null;

    Cursor cursor = getReadableDatabase().rawQuery("Select _LESSON_PROBLEM from "+SQLiteHelper.TABLE_NAME_STUDENTSPECIFICLESSON+" Where "
            +SQLiteHelper._viewID +"='"+_id+"'",null);


    if (cursor != null) {
        cursor.moveToFirst();
    }
    if (cursor == null) {
    } else if (cursor.moveToFirst()) {
        do {
            id = String.valueOf(cursor.getInt(cursor.getColumnIndex("_LESSON_PROBLEM")));

        } while (cursor.moveToNext());
        cursor.close();
    } else {

    }
    return id;

}

从我的Adapter电话:

String lessOnProblem= helper.getLessonProblem(viewId);

以上String lessonProblem将返回0.

我检查了我的数据库并正确插入了所有数据,这是我的数据库的图像:

我不明白为什么它显然返回0时显然不是,有人可以帮我指出问题可能是什么？

马虎的错误.你应该使用getString()而不是getInt().

以String形式返回所请求列的值.结果以及当列值为null或列类型不是字符串类型时此方法是否抛出异常是实现定义的.

最后

cursor.getString(cursor.getColumnIndex("_LESSON_PROBLEM"));

id = String.valueOf(cursor.getInt(cursor.getColumnIndex("_LESSON_PROBLEM")));

您将字符串值作为int获取,然后将其转换为字符串.把它改成类似的东西

id = cursor.getString(cursor.getColumnIndex("_LESSON_PROBLEM"));