我看不到添加到我的 ListView 的第一个项目
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【中文标题】我看不到添加到我的 ListView 的第一个项目【英文标题】：I can't see the first item added to my ListView 【发布时间】：2018-06-09 02:48:36 【问题描述】：我有一个使用 OpenSQLiteHelper 制作的数据库，我从中获取一串名称并将其添加到 ListView。我永远看不到数据库中的第一项，而 ListView 仅从数据库中的第二项开始。
我正在从我的数据库中的 COL_2（名称）获取字符串。
ListView 类
public class Castawaylist extends AppCompatActivity
private static final String TAG = "CastList";
myDbAdapter mydb;
private ListView castview;
@Override
protected void onCreate(Bundle savedInstanceState)
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_castawaylist);
Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
setSupportActionBar(toolbar);
FloatingActionButton fab = (FloatingActionButton) findViewById(R.id.fab);
fab.setOnClickListener(new View.OnClickListener()
@Override
public void onClick(View view)
Snackbar.make(view, "Replace with your own action", Snackbar.LENGTH_LONG)
.setAction("Action", null).show();
);
castview = (ListView) findViewById(R.id.listView);
mydb = new myDbAdapter(this);
populateCastList();
String score;
private void populateCastList()
//get the data and append to the list
Log.d(TAG, "populateCastList: Displaying List of Castaways.");
Cursor data = mydb.getData();
//int save = data.getInt(2);
ArrayList<String> listData = new ArrayList<>();
while(data.moveToNext())
listData.add(data.getString(1) +
" " + data.getInt(2)); //get value from database at column 1, then add it to array list
ListAdapter adapter = new ArrayAdapter<>(this, android.R.layout.simple_list_item_1, listData); //Creates the list adapter and set it
castview.setAdapter(adapter);
castview.setOnItemClickListener(new AdapterView.OnItemClickListener()
@Override
public void onItemClick(AdapterView<?> adapterView, View view, int i, long l)
String nametest = "";
int u = 0;
while(!adapterView.getItemAtPosition(i).toString().substring(u,u+1).equals(" "))
nametest = nametest + adapterView.getItemAtPosition(i).toString().substring(u, u+1);
u = u + 1;
Log.d(TAG, "onItemClick: You Clicked on " + nametest);
Cursor data = mydb.getItemID(nametest); //Get the ID associated with that name
int itemID = -1;
while(data.moveToNext())
itemID = data.getInt(0);
if(itemID > -1)
Log.d(TAG, "onItemClick: The ID is: " + itemID);
Intent editCastaway = new Intent(Castawaylist.this, EditCastaway.class);
editCastaway.putExtra("id", itemID);
editCastaway.putExtra("name", nametest);
//editCastaway.putExtra("score", data.getInt(2));
startActivity(editCastaway);
else
toastText("No ID associated with that name");
);
private void toastText(String text)
Toast.makeText(this,text, Toast.LENGTH_SHORT).show();
ListView XML 代码
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical"
android:layout_
android:layout_>
<ListView
android:layout_
android:layout_
android:id="@+id/listView"/>
</LinearLayout>
数据库助手类
public class myDbAdapter extends SQLiteOpenHelper
public static final String DATABASE_NAME = "Castaways.db";
public static final String TABLE_NAME = "Survivor_Table";
public static final String COL_1 = "ID";
public static final String COL_2 = "NAME";
public static final String COL_3 = "MARK";
public myDbAdapter(Context context)
super(context, DATABASE_NAME, null, 2);
@Override
public void onCreate(SQLiteDatabase db)
db.execSQL("create table " + TABLE_NAME +" (ID INTEGER PRIMARY KEY AUTOINCREMENT,NAME TEXT,MARK INTEGER)");
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion)
db.execSQL("DROP TABLE IF EXISTS "+TABLE_NAME);
onCreate(db);
public boolean addData(String name,int score)
SQLiteDatabase db = this.getWritableDatabase();
ContentValues contentValues = new ContentValues();
contentValues.put(COL_2,name);
contentValues.put(COL_3,score);
long result = db.insert(TABLE_NAME,null ,contentValues);
if(result == -1)
return false;
else
return true;
public Cursor getData()
SQLiteDatabase db = this.getWritableDatabase();
Cursor res = db.rawQuery("select * from "+TABLE_NAME,null);
return res;
public Cursor getItemID(String name)
SQLiteDatabase db = this.getWritableDatabase();
String query = "SELECT " + COL_1 + " FROM " + TABLE_NAME +
" WHERE " + COL_2 + " = '" + name + "'";
Cursor data = db.rawQuery(query, null);
return data;
public boolean updateData(String id,String name,String surname,int score)
SQLiteDatabase db = this.getWritableDatabase();
ContentValues contentValues = new ContentValues();
contentValues.put(COL_1,id);
contentValues.put(COL_2,name);
contentValues.put(COL_3,score);
db.update(TABLE_NAME, contentValues, "ID = ?",new String[] id );
return true;
public void updateName(String newName, int id, String oldName)
SQLiteDatabase db = this.getWritableDatabase();
String query = "UPDATE " + TABLE_NAME + " SET " + COL_2 +
" = '" + newName + "' WHERE " + COL_1 + " = '" + id + "'" +
" AND " + COL_2 + " = '" + oldName + "'";
Log.d(TAG, "updateName: query: " + query);
Log.d(TAG, "updateName: Setting name to " + newName);
db.execSQL(query);
public Integer deleteData(String id)
SQLiteDatabase db = this.getWritableDatabase();
return db.delete(TABLE_NAME, "ID = ?",new String[] id);
public void deleteName(int id, String name)
SQLiteDatabase db = this.getWritableDatabase();
String query = "DELETE FROM " + TABLE_NAME + " WHERE "
+ COL_1 + " = '" + id + "'" +
" AND " + COL_2 + " = '" + name + "'";
Log.d(TAG, "deleteName: query: " + query);
Log.d(TAG, "deleteName: Deleting " + name + " from database.");
db.execSQL(query);
public void updateScore(int newScore, int id, int oldScore)
SQLiteDatabase db = this.getWritableDatabase();
String query = "UPDATE " + TABLE_NAME + " SET " + COL_3 +
" = '" + newScore + "' WHERE " + COL_1 + " = '" + id + "'" +
" AND " + COL_3 + " = '" + oldScore + "'";
Log.d(TAG, "updateName: query: " + query);
Log.d(TAG, "updateName: Setting name to " + newScore);
db.execSQL(query);
public void clearData()
SQLiteDatabase db = this.getWritableDatabase();
db.delete(TABLE_NAME,null,null);
db.close();
感谢您的帮助！
【问题讨论】：
你怎么知道有一行你没有看到？ 【参考方案1】：不确定，但问题似乎在这里： 而（数据。moveToNext（）） 可能是 moveToNext() 在离开前一个项目后移动到下一个项目时触发，换句话说，它将在离开项目 #1 后第一次触发） 尝试使用索引循环代替。即for (int i=0; ....
【讨论】：
【参考方案2】：在这段代码中：
while(data.moveToNext())
itemID = data.getInt(0);
它将触发 moveToNext() 函数，然后执行 while。所以你可以使用 do while 来解决它
do
itemID = data.getInt(0);
while(data.moveToNext())
【讨论】：
我也试过了，但出现运行时异常原因：android.database.CursorIndexOutOfBoundsException: Index -1 requested, with a size of 6【参考方案3】：在我使用光标的研究中，您的初始设置是正确的
while(data.moveToNext())...
光标在行之前开始，初始调用 moveToNext() 将其移动到第一行。
一个明显的问题是您没有关闭光标。
while(data.moveToNext())
listData.add(data.getString(1) +
" " + data.getInt(2)); //get value from database at column 1, then add it to array list
// Close the Cursor here and again after the second one you create
data.close();
我看到的第二个潜在问题是您没有在 onClick 中访问正确的值
// int i might not be 0 at some point
@Override
public void onItemClick(AdapterView<?> adapterView, View view, int i, long l)
i--;// maybe try subtracting 1 from i. Alternative is to do that when i is being used
String nametest = "";
int u = 0;
// or subtract here
while(!adapterView.getItemAtPosition(i-1).toString().substring(u,u+1).equals(" "))
nametest = nametest + adapterView.getItemAtPosition(i-1).toString().substring(u, u+1);
u = u + 1;
【讨论】：
我试过这个并得到一个致命的运行时异常原因：android.database.CursorIndexOutOfBoundsException: Index -1 requested, with a size of 6 在文档中，您可以查找几种不同类型的内容，但如果您遇到 IndexOutOfBoundsException，您可能需要尝试使用其他内容，例如 !data.isLast()。或者您可以在进入while循环之前确保光标位于第一个位置 data.moveToFirst() 嗨，我也试过了，但它仍然给我异常错误，但我注意到我的数据库中有 6 项我正试图放入列表中。当我再添加一个错误更改为 Caused by: android.database.CursorIndexOutOfBoundsException: Index -1 requested, with a size of 7. 它实际上是添加所有项目还是不添加？使用带有断点的调试检查 listData.add(); 您可能需要尝试使用计数器而不是位置。以上是关于我看不到添加到我的 ListView 的第一个项目的主要内容，如果未能解决你的问题，请参考以下文章
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