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我已經在SQLite的android中創建了兩個表,但是每當我運行這段代碼時它說沒有找到那個名字的表,我已經在很多設備上嘗試了它,但我仍然無法找到錯誤下面的代碼。SQLite Android - 表未找到
public void onOpen(SQLiteDatabase db) {
super.onOpen(db);
}
@Override
public void onCreate(SQLiteDatabase db) {
db.execSQL("create table peoples" +
"(personid integer primary key autoincrement," +
" pname text," +
" pvehicle text," +
" ptime text," +
" pdate text," +
" startpoint text," +
" endpoint text)");
}
@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
db.execSQL("DROP TABLE IF EXISTS peoples");
}
public boolean insertPeoples(String pname,String pvehicle, String ptime, String pdate, String start, String end)
{
try {
SQLiteDatabase db = this.getWritableDatabase();
ContentValues values = new ContentValues();
values.put("pname",pname);
values.put("pvehicle",pvehicle);
values.put("ptime",ptime);
values.put("pdate",pdate);
values.put("startpoint",start);
values.put("endpoint",end);
db.insert("peoples",null,values);
}
catch (SQLiteException ex)
{
Log.i("MyLog",ex.getMessage());
}
return true;
}
public Cursor getAllpeoples()
{
Cursor res = null;
try {
SQLiteDatabase database = this.getReadableDatabase();
res = database.rawQuery("select * from peoples",null);
}
catch (SQLiteException ex)
{
Log.e("MyLog",ex.getMessage());
}
return res;
}
public void deletePeoples()
{
SQLiteDatabase db = this.getWritableDatabase();
db.execSQL("delete from peoples");
}
存在具有類似的代碼另一個表,並工作正常,但第二個表由另一個類處理,並工作正常,但此代碼顯示錯誤,沒有與名人們發現並表,所以它返回null指針異常,而我試圖獲取記錄。
不,這不是問題 –