我创建DbManger类来处理Sqlite数据库操作,我尝试在数据库中插入值,它给出了错误
E / SQLiteDatabase(6729):android.database.sqlite.SQLiteException:near“CREATE”:语法错误(代码1):,同时编译:INSERT INTO CREATE TABLE关系(_id INTEGER PRIMARY KEY AUTOINCREMENT,name TEXT NOT NULL ,发送电子邮件TEXT NOT NULL,年龄TEXT NOT NULL,位置TEXT NOT NULL);(电子邮件,姓名,年龄,位置)VALUES(?,?,?,?)
DbManager Class
public class DBManager {
int rowsAffected;
// Database and version
private static final String DB_NAME = "familyHistory";
private static final int DB_VERSION = 1;
private static final String TABLE_INFO = "familyTable";
private static final String _ROWID = "_id";
private static final String _NAME = "name";
private static final String _EMAIL = "email";
private static final String _AGE = "age";
private static final String _LOCATIONs = "location";
private static final String CREATE_PRO = "CREATE TABLE " + TABLE_INFO + "(" +
_ROWID + " INTEGER PRIMARY KEY AUTOINCREMENT, " +
_NAME + " TEXT NOT NULL, " +
_EMAIL + " TEXT NOT NULL, " +
_AGE + " TEXT NOT NULL, " +
_LOCATIONs + " TEXT NOT NULL);";
private final Context ourContext;
private static DBHelper ourDBHelper;
private SQLiteDatabase ourDatabase;
public DBManager(Context ctx) {
this.ourContext = ctx;
Log.i("db", "DBManager(Context ctx)");
}
private static class DBHelper extends SQLiteOpenHelper {
public DBHelper(Context context) {
// SQLiteOpenHelper Constructor Creating database
super(context, DB_NAME, null, DB_VERSION);
Log.i("db", "DBHelper(Context context)");
}
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL(CREATE_PRO);
Log.e(">>>>>", CREATE_PRO);
// db.execSQL(CREATE_TRVLBOK);
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVer, int newVer) {
Log.w("Nomad", "Upgrading database from version " + oldVer + " to " + newVer
+ ", which will destroy all old data");
db.execSQL("DROP TABLE IF EXISTS " + TABLE_INFO);
// db.execSQL("DROP TABLE IF EXISTS " + TABLE_TRAVELBOK);
onCreate(db);
Log.i("db", "onUpgrade)");
}
}
public DBManager open() throws SQLException {
ourDBHelper = new DBHelper(ourContext);
ourDatabase = ourDBHelper.getWritableDatabase();
Log.i("db", "open()");
return this;
}
public void close() {
ourDBHelper.close();
Log.i("db", "close()");
}
public long insertFamRec(String UserName, String Email, String age, String location) {
ContentValues cv = new ContentValues();
cv.put(_NAME, UserName);
cv.put(_EMAIL, Email);
cv.put(_AGE, age);
cv.put(_LOCATIONs, location);
Log.i("db", "insertLocRec");
return ourDatabase.insert(CREATE_PRO, null, cv);
}
}
从插入类我添加了这些行但仍然是错误
DBManager db = new DBManager(NewRelation.this);
db.open();
long i = db.insertFamRec("strfullName", "stremail", "strage", "strlocation");
db.close();
if (i > 0) {
ShowMessage.Message(NewRelation.this, "data is added");
} else {
Toast.makeText(getApplicationContext(), "data is failed", Toast.LENGTH_SHORT).show();
}
如果有人放弃了对问题的评价,请提及原因。
答案 0 :(得分:1)
问题似乎是你调用ourDatabase.insert
的地方,你在表名的位置传递查询字符串。您希望通过这一行实现什么目标:return ourDatabase.insert(CREATE_PRO, null, cv);
?
将行更改为:
return ourDatabase.insert(TABLE_INFO, null, cv);
注意,我传递CREATE_PRO
而不是TABLE_INFO
,这是您要插入的表的名称。
我希望这可以解决你的问题。如果有帮助,请告诉我。