首先让我解释一下如何从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);
}

然后在sqlite中这样插入到表中:
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);
}

我从这样的活动中插入:
sqLiteHelper.insertLesson(id,etLessonTitle.getText().toString(),currentDateandTime,etLessonProbStu.getText().toString());

您会注意到我已经创建了_viewID INTEGER PRIMARY KEY AUTOINCREMENT我正在使用它来获取适配器中的特定viewholder。
我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;
}

这很好地工作,并按预期正确返回_viewID
然后,我尝试通过执行以下操作从sqlite获取_viewId
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_LESSON_PROBLEM中,我打电话给:
String lessonProblem = helper.getLessonProblem(viewId);

上述Adapter将返回String lessonProblem
我已经检查了我的数据库,所有数据都正确插入,下面是我的数据库的图像:
android - SQlite返回0-LMLPHP
我不明白为什么它会返回0,但显然不是,有人能帮我指出问题所在吗?

最佳答案

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

得到一个字符串值作为int,然后将其转换为字符串。把它改成
id = cursor.getString(cursor.getColumnIndex("_LESSON_PROBLEM"));

09-11 19:51
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