我正在制作一个我希望保存用户联系人详细信息的应用程序。但每当我尝试插入或选择某些值时,我都会收到错误。
DataBasae代码
public class ContactsDatabase extends SQLiteOpenHelper {
private static final int dbVersion = 1;
private static final String dbName = "HSsuraksha";
private static final String tableName = "contactsDetails";
private static final String contactId = "contactId";
private static final String groupId = "groupId";
private static final String groupGroupId = "groupId";
private static final String groupTable = "groupDetails";
private static final String contactName = "contactName";
private static final String contactNumber = "contactNumber";
private static final String createTable = "Create Table " + tableName + "(" + contactId + " Integer Primary Key AutoIncrement," + groupId + " Text," + contactName + " Text," + contactNumber + " Text" + ");";
public ContactsDatabase(Context context) {
super(context, dbName, null, dbVersion);
}
@Override
public void onCreate(SQLiteDatabase sqLiteDatabase) {
sqLiteDatabase.execSQL(createTable);
}
@Override
public void onUpgrade(SQLiteDatabase sqLiteDatabase, int i, int i2) {
}
public void insertContacts(ContactModel contactModel, String id, ArrayList<ContactModel> contactModelArrayList) {
SQLiteDatabase database = getWritableDatabase();
database.beginTransaction();
ContentValues contentValues = new ContentValues();
for (int i = 0; i < contactModelArrayList.size(); i++) {
contentValues.put(contactName, contactModelArrayList.get(i).getContactName());
contentValues.put(contactNumber, contactModelArrayList.get(i).getContactNumber());
contentValues.put(groupId, id);
if (contentValues != null) {
Long value = database.insert(tableName, id, contentValues);
}
}
database.setTransactionSuccessful();
database.close();
}
public void selectContacts(String id) {
String query = "Select * From " + tableName + " where " + groupId + "=?";
SQLiteDatabase database = getWritableDatabase();
Cursor cursor = database.rawQuery(query, new String[]{id});
while (cursor.moveToNext()) {
cursor.getString(cursor.getColumnIndexOrThrow(contactName));
}
cursor.close();
database.close();
}
}
logcat的
08-01 18:53:57.820 672-672/example.com.pocketdocs E/SQLiteLog﹕ (1) no such table: contactsDetails
08-01 18:53:57.830 672-672/example.com.pocketdocs E/SQLiteDatabase﹕ Error inserting groupId=2 contactNumber=+91 97 69 512114 contactName=Aaaaaaa
android.database.sqlite.SQLiteException: no such table: contactsDetails (code 1): , while compiling: INSERT INTO contactsDetails(groupId,contactNumber,contactName) VALUES (?,?,?)
at android.database.sqlite.SQLiteConnection.nativePrepareStatement(Native Method)
at android.database.sqlite.SQLiteConnection.acquirePreparedStatement(SQLiteConnection.java:893)
at android.database.sqlite.SQLiteConnection.prepare(SQLiteConnection.java:504)
at android.database.sqlite.SQLiteSession.prepare(SQLiteSession.java:588)
at android.database.sqlite.SQLiteProgram.<init>(SQLiteProgram.java:58)
at android.database.sqlite.SQLiteStatement.<init>(SQLiteStatement.java:31)
at android.database.sqlite.SQLiteDatabase.insertWithOnConflict(SQLiteDatabase.java:1475)
at android.database.sqlite.SQLiteDatabase.insert(SQLiteDatabase.java:1347)
at example.com.pocketdocs.DataBase.ContactsDatabase.insertContacts(ContactsDatabase.java:54)
at example.com.pocketdocs.Group.CreateNewGroup.onClick(CreateNewGroup.java:104)
at android.view.View.performClick(View.java:4147)
at android.view.View$PerformClick.run(View.java:17161)
at android.os.Handler.handleCallback(Handler.java:615)
at android.os.Handler.dispatchMessage(Handler.java:92)
at android.os.Looper.loop(Looper.java:213)
at android.app.ActivityThread.main(ActivityThread.java:4787)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:511)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:789)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:556)
at dalvik.system.NativeStart.main(Native Method)
我尝试解决错误但没有成功。我的代码是什么问题???
答案 0 :(得分:9)
我有另一个表groupInfo与相同的数据库名称,所以它的问题??
这是一个问题。这是发生的事情:
访问具有相同数据库文件的第一个sqlite打开助手。如果数据库文件不存在,则调用onCreate()
回调,以便您可以设置数据库文件。
访问具有相同数据库文件的其他sqlite打开助手。具有给定名称的数据库文件已存在且版本正确,因此不会调用onCreate()
或onUpgrade()
。而是刚刚打开文件。
解决方案:每个数据库文件只使用一个sqlite open helper。将两个表的创建语句放在同一帮助器onCreate()
方法中。
同时卸载您的应用,以便删除仅包含其他表的旧数据库文件。
请参阅链接问题When is SQLiteOpenHelper onCreate() / onUpgrade() run?以了解有关sqlite open helper生命周期回调的更多信息。
答案 1 :(得分:0)
将private static final String dbName = "HSsuraksha";
更改为private static final String dbName = "HSsuraksha.db";
你创建了数据库,但没有扩展名,所以像HSsuraksha.db
答案 2 :(得分:0)
您的表格未创建,因此系统无法插入数据。
android.database.sqlite.SQLiteException:没有这样的表: contactsDetails(代码1):
我建议你查看你的SQL请求然后使用这段代码:
private static final String DATABASE_NAME = "YOURDATABASE.db";
public DBHelper(Context context, String tableName) {
super(context, DATABASE_NAME, null, DATABASE_VERSION);
mTableName = tableName;
}
@Override
public void onCreate(SQLiteDatabase database) {
Log.d(TAG, "Create application database");
database.execSQL(createTable(mTableName));
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
Log.w(TAG, "Upgrading database from version " + oldVersion + " to " + newVersion + ", which will destroy all old data");
db.execSQL("DROP TABLE IF EXISTS " + mTableName);
onCreate(db);
}
private String createTable(String tableName) {
return createTable;
}
答案 3 :(得分:0)