android SQLiteConstraintException唯一约束失败

如何解决此异常pleaz help me我已将项目添加到网格视图，但当我单击网格视图的最后一个项目时，此异常显示该项目已正确删除，但此异常引发了代码中的错误是什么

public class MyDataBase extends SQLiteOpenHelper {
private static final int DATABASE_VERSION = 1;
// Database Name
private static final String DATABASE_NAME = "contactsManager";
// Contacts table name
private static final String TABLE_CONTACTS = "contacts";
private static final String TABLE_DELETED = "deleted";
// Contacts Table Columns names
private static final String KEY_ID = "id";
final String KEY_NAME = "name";
private static final String KEY_dID = "did";
final String KEY_dNAME = "dname";
public MyDataBase(Context context) {
    super(context, DATABASE_NAME, null, DATABASE_VERSION);
}

@Override
public void onCreate(SQLiteDatabase db) {
    String CREATE_CONTACTS_TABLE = "CREATE TABLE " + TABLE_CONTACTS + "("
            + KEY_ID + " INTEGER PRIMARY KEY," + KEY_NAME + " TEXT UNIQUE"
            + ")";
    String CREATE_DELETED_TABLE = "CREATE TABLE " + TABLE_DELETED + "("
            + KEY_dID + " INTEGER PRIMARY KEY," + KEY_dNAME + " TEXT UNIQUE"
            + ")";

    db.execSQL(CREATE_CONTACTS_TABLE);
    db.execSQL(CREATE_DELETED_TABLE);
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
}
public void AddnewContact(photos todo)// this method to add new contact
{
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(KEY_NAME, todo.getImage());
    db.insert(TABLE_CONTACTS, null, values);
    db.close();
}
public void AddnewContactToDeletedTable(photos todo)// this method to add new contact
{
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues values = new ContentValues();
    values.put(KEY_dNAME, todo.getImage());
    db.insert(TABLE_DELETED, null, values);
    db.close();
}
public ArrayList<photos> getallcontacts() {
    ArrayList<photos> contactList = new ArrayList<photos>();
    String selectQuery = "SELECT  * FROM " + TABLE_CONTACTS;
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
    if (cursor.moveToFirst()) {
        do {
            photos contact = new photos();
            contact.setId(Integer.parseInt(cursor.getString(0)));
            contact.setImage(cursor.getString(1));
            // Adding contact to list
            contactList.add(contact);
        } while (cursor.moveToNext());
    }
    // return contact list
    return contactList;
}
public void deleteContact(photos contact) {
    int id = contact.getId();
    SQLiteDatabase db = this.getWritableDatabase();
    db.delete(TABLE_CONTACTS, KEY_ID
            + " = " + id, null);
    db.close();
}

---

我认为您插入了相同的值，这就是为什么抛出此错误的原因，另一种解决方案是您可以替换查询

请尝试以下代码：

public void AddnewContact(photos todo)// this method to add new contact
{
   SQLiteDatabase db = this.getWritableDatabase();
   ContentValues values = new ContentValues();
   values.put(KEY_NAME, todo.getImage());
   db.replace(TABLE_CONTACTS, null, values); // change this line to your code
   db.close();
}

替换查询可以处理已存在的数据

public void AddnewContact(photos todo)// this method to add new contact
{
   SQLiteDatabase db = this.getWritableDatabase();
   ContentValues values = new ContentValues();
   values.put(KEY_NAME, todo.getImage());
   db.replace(TABLE_CONTACTS, null, values); // change this line to your code
   db.close();
}