Android Sqlite查询返回null
我在Android中有一个静态sqlite数据库。函数接受int类型的输入并对数据库进行查询。当输入值达到97500时,它可以正常工作,但是如果我输入任何更大的值,则会出现两种情况之一Android Sqlite查询返回null,android,android-sqlite,Android,Android Sqlite,我在Android中有一个静态sqlite数据库。函数接受int类型的输入并对数据库进行查询。当输入值达到97500时,它可以正常工作,但是如果我输入任何更大的值,则会出现两种情况之一 如果输入为98000-99500,则返回null 如果输入大于100000,则返回错误的数据 以下是出现故障的函数: Budget getBudget(int income,String name) { Budget B=null; Log.d("DB", String.valueOf(i
Budget getBudget(int income,String name)
{
Budget B=null;
Log.d("DB", String.valueOf(income));
try{
SQLiteDatabase db=this.getReadableDatabase();
Cursor cur=db.rawQuery("SELECT * FROM "+BudgetTable+" WHERE "+Low+"<=? AND "+High+">=?", new String[]{String.valueOf(income),String.valueOf(income)});
Log.d("DB", String.valueOf(cur.getCount()));
if(cur.getCount()!=0)
{
Log.d("DB", "Cursor not empty");
cur.moveToFirst();
B=new Budget(0, income,cur.getInt(cur.getColumnIndex(MortgageRent)),cur.getInt(cur.getColumnIndex(Utilities)) ,
cur.getInt(cur.getColumnIndex(LightnPower)), cur.getInt(cur.getColumnIndex(PhonenInternet)),
cur.getInt(cur.getColumnIndex(HomeMaintenance)), cur.getInt(cur.getColumnIndex(HomeCleaning)),
cur.getInt(cur.getColumnIndex(Groceries)), cur.getInt(cur.getColumnIndex(Clothing)),0,
cur.getInt(cur.getColumnIndex(PersonalGrooming)), cur.getInt(cur.getColumnIndex(MedicalnPharmacy)),
cur.getInt(cur.getColumnIndex(HealthInsurance)), cur.getInt(cur.getColumnIndex(LifeInsurance)),
cur.getInt(cur.getColumnIndex(HomeInsurance)), cur.getInt(cur.getColumnIndex(Accounting)),
cur.getInt(cur.getColumnIndex(BankFees)), cur.getInt(cur.getColumnIndex(Fuel)),
cur.getInt(cur.getColumnIndex(ServicenRepairs)), cur.getInt(cur.getColumnIndex(GovernmentCharges)),
cur.getInt(cur.getColumnIndex(CarInsurance)),0, cur.getInt(cur.getColumnIndex(PublicTransport)),
cur.getInt(cur.getColumnIndex(Entertainment)), cur.getInt(cur.getColumnIndex(SportsnGym)),
cur.getInt(cur.getColumnIndex(EatOut)), cur.getInt(cur.getColumnIndex(Alcohol)),
cur.getInt(cur.getColumnIndex(Gifts)), cur.getInt(cur.getColumnIndex(Holidays)),
cur.getInt(cur.getColumnIndex(NewspapernMagazine)), cur.getInt(cur.getColumnIndex(Others)), 0, 0, name);
Log.d("DB", String.valueOf(cur.getInt(cur.getColumnIndex(IncomeLevel))));
cur.close();
db.close();
}
}catch(Exception ex)
{
Log.d("DB", ex.getMessage());
}
return B;
}
Budget getBudget(整数收入,字符串名称)
{
预算B=空;
Log.d(“DB”,String.valueOf(income));
试一试{
SQLiteDatabase db=this.getReadableDatabase();
游标cur=db.rawQuery(“从“+BudgetTable+”中选择*,其中“+Low+”=?”,新字符串[]{String.valueOf(income),String.valueOf(income)});
Log.d(“DB”,String.valueOf(cur.getCount());
如果(cur.getCount()!=0)
{
Log.d(“DB”,“光标不为空”);
cur.moveToFirst();
B=新预算(0,收入,当前getInt(当前getColumnIndex(MortgageRent)),当前getInt(当前getColumnIndex(Utilities)),
cur.getInt(cur.getColumnIndex(LightnPower)),cur.getInt(cur.getColumnIndex(PhonenInternet)),
cur.getInt(cur.getColumnIndex(HomeMaintenance)),cur.getInt(cur.getColumnIndex(HomeCleaning)),
当前getInt(当前getColumnIndex(杂货)),当前getInt(当前getColumnIndex(服装)),0,
cur.getInt(cur.getColumnIndex(个性化修饰)),cur.getInt(cur.getColumnIndex(MedicalPharmacy)),
当前getInt(当前getColumnIndex(健康保险)),当前getInt(当前getColumnIndex(人寿保险)),
当前getInt(当前getColumnIndex(家庭保险)),当前getInt(当前getColumnIndex(会计)),
当前getInt(当前getColumnIndex(银行费用)),当前getInt(当前getColumnIndex(燃料)),
cur.getInt(cur.getColumnIndex(ServicenRepairs)),cur.getInt(cur.getColumnIndex(GovernmentCharges)),
cur.getInt(cur.getColumnIndex(CarInsurance)),0,cur.getInt(cur.getColumnIndex(PublicTransport)),
cur.getInt(cur.getColumnIndex(娱乐)),cur.getInt(cur.getColumnIndex(SportsnGym)),
cur.getInt(cur.getColumnIndex(EatOut)),cur.getInt(cur.getColumnIndex(酒精)),
当前getInt(当前getColumnIndex(礼品)),当前getInt(当前getColumnIndex(假日)),
cur.getInt(cur.getColumnIndex(NewspapernMagazine)),cur.getInt(cur.getColumnIndex(其他)),0,0,name);
Log.d(“DB”,String.valueOf(cur.getInt(cur.getColumnIndex(IncomeLevel)));
cur.close();
db.close();
}
}捕获(例外情况除外)
{
Log.d(“DB”,例如getMessage());
}
返回B;
}
下面是数据库中数据的截图…我不明白为什么它不工作。
由于我看不到您的数据库,我相信您的问题的根源是这一行:
Cursor cur=db.rawQuery("SELECT * FROM "+BudgetTable+" WHERE "+Low+"<=? AND "+High+">=?", new String[]{String.valueOf(income),String.valueOf(income)});
您不应该将数字作为字符串传递,因为字符串具有不同的比较规则,例如,字符串“2000”将被视为大于字符串“102500” 他们在javadoc文档中对此提出警告: selectionArgs您可以在查询中的where子句中包含?s,该子句将被selectionArgs中的值替换。这些值将被绑定为字符串 您应该按照以下方式重写查询:
Cursor cur=db.rawQuery("SELECT * FROM "+BudgetTable+
" WHERE "+Low+"<=" + income + " AND "+High+">=" + income);
cursorcur=db.rawQuery(“选择*自”+BudgetTable+
“其中”+低+“=”+收入);
此外,这是一个非常常见的问题,有些人遇到了类似的问题:否查询是我想要的,并且返回的输入数据少于98000的正确数据。我还附上了数据库的截图。如果收入=98500,则应选择rowid=17,低=97500,高=102500,这显然验证了97500=98500。但是它返回空值。谢谢你的建议。正如您所说,我已经编写了查询,但现在它总是返回null。@Mehedi发布了一个创建BudgetTable的代码。我需要验证一些字段是否为整数类型。问题已经解决,问题相当愚蠢…当我将数据从csv导入静态数据库时,一些字段已损坏,这些字段正在创建错误…我已修复csv并再次导入数据,一切正常。谢谢你的帮助。
Cursor cur=db.rawQuery("SELECT * FROM "+BudgetTable+
" WHERE "+Low+"<=" + income + " AND "+High+">=" + income);