7

我正在尝试从数据库的某一行中获取文本。为此，我做了这个功能

public String getTranslation(long rowId) throws SQLException {    
    String str = null;
    Cursor mCursor = db.query(
        true, TABLE_LANGUAGE_FR, new String[] {
            KEY_ID_INTERFACE_TEXT, KEY_ITEM_NAME,KEY_FORM_LABEL_ID, KEY_FORM_RANK
        }, 
        KEY_FORM_LABEL_ID + "=" + rowId, null, null, null, null, null
    );

    if (mCursor != null) {
        mCursor.moveToFirst();
        str = mCursor.getString(mCursor.getColumnIndex(KEY_ITEM_NAME));
    }
    return str;
}

我这样称呼它：

db = new DbAdapter(this);
db.createDatabase(); 
db.openDataBase();
String str = db.getTranslation(1706);

我的桌子看起来像这样：

在此处输入图像描述

我收到此错误：

09:32:08.965    307 com.Orange  ERROR   AndroidRuntime  Caused by: android.database.CursorIndexOutOfBoundsException: Index 0 requested, with a size of 0
09:32:08.965    307 com.Orange  ERROR   AndroidRuntime      at android.database.AbstractCursor.checkPosition(AbstractCursor.java:580)
09:32:08.965    307 com.Orange  ERROR   AndroidRuntime      at android.database.AbstractWindowedCursor.checkPosition(AbstractWindowedCursor.java:214)
09:32:08.965    307 com.Orange  ERROR   AndroidRuntime      at android.database.AbstractWindowedCursor.getString(AbstractWindowedCursor.java:41)
09:32:08.965    307 com.Orange  ERROR   AndroidRuntime      at com.Orange.Database.DbAdapter.getTranslation(DbAdapter.java:141)
09:32:08.965    307 com.Orange  ERROR   AndroidRuntime      at com.Orange.Authentication.onCreate(Authentication.java:32)

为什么我会收到此错误？我在这里做错了什么？

提前致谢。

4

3 回答 3

25

It would seem like you have an emtpy cursor. Instead of checking whether it is null try checkcing 

if(mCursor.getCount() > 0)
{
  //do stuff
}

try removing your where condition to make sure you actually get data when running the select statement. If you still get nothing you might have a typo in your tablenames, or failed to connect/create the database/tables.

Cursor mCursor = db.query("sys_interface_text", new String[] {
       "id_interface_text", "item_name","form_label_id", "form_rank"}, 
               "form_label_id = 1706", null, null, null,
       null);
于 2012-06-29T07:08:41.500 回答
1

尝试使用

while(cursor.moveToNext())
{
   if(cursor.isFirst())
   {
      //Your code goes here in your case
      return mCursor.getString(mCursor.getColumnIndex(KEY_ITEM_NAME));
   }
}
finaly
{
 if(mCursor!= null)
 {
   mCursor.close();
 }
}
于 2012-06-29T08:26:15.770 回答
0

尝试更改此行

Cursor mCursor = db.query(true, TABLE_LANGUAGE_FR, new String[] {
           KEY_ID_INTERFACE_TEXT, KEY_ITEM_NAME,KEY_FORM_LABEL_ID, KEY_FORM_RANK}, 
                   KEY_FORM_LABEL_ID + "=" + rowId, null, null, null,
           null, null);

经过

Cursor mCursor = db.query(TABLE_LANGUAGE_FR, new String[] {
           KEY_ID_INTERFACE_TEXT, KEY_ITEM_NAME,KEY_FORM_LABEL_ID, KEY_FORM_RANK}, 
                   KEY_FORM_LABEL_ID + "=" + rowId, null, null, null,
           null);

并将其添加到 if 条件中

 return str;
于 2012-06-29T07:10:56.277 回答