在iPhone中插入特殊字符的數據庫問題？

Question

我是sqlite數據庫和iPhone開發的新手

我將數據插入數據庫

am insert description string is like dz 
• 500 grams Chicken breasts
• 6-8 nos. Green chillies
• 2 tblsp Coriander leaves
• 2 tblsp Mint leaves
• 12 nos. Garlic
• 8-10 nos. Almonds
• 3/4 cup Fresh cream
• 1 inch Ginger
• 1 tblsp Lemon juice
• Oil
• Salt
this is code for inserting 

-(void)insertDataToFavourites:(int)pageid111 :(NSString *)description
{
    NSLog(@"%@",description);
    const char *sqlStatement = "INSERT INTO AddFavoritenew (Id,Description) VALUES (?,?)";
    sqlite3_stmt *compiled_statement;
    if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiled_statement, NULL) == SQLITE_OK)
    {
        sqlite3_bind_int(compiled_statement, 1, pageid111);
        sqlite3_bind_text(compiled_statement, 2, [description UTF8String] , -1, SQLITE_TRANSIENT);
        [ArrAddData addObject:[NSString stringWithFormat:@"%d", pageid111]];
        NSLog(@"ArrAddData record is:--> %@",ArrAddData);
    }

    if(sqlite3_step(compiled_statement) != SQLITE_DONE )
    {
        //        NSLog( @"Error: %s", sqlite3_errmsg(database) );
        UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Alert!"
                                                        message:@"alerady you added this sms to favorite list."
                                                       delegate:self
                                              cancelButtonTitle:@"Ok"
                                              otherButtonTitles:nil, nil];
        [alert show];
        [alert release];
    }
    else
    {
        //        NSLog( @"Insert into row id = %lld", sqlite3_last_insert_rowid(database));
        UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Alert!"
                                                        message:@"Data saved to Favorite."
                                                       delegate:self
                                              cancelButtonTitle:@"Ok"
                                              otherButtonTitles:nil, nil];
        [alert show];
        [alert release];
    }

    sqlite3_finalize(compiled_statement);    
}

正在將上述數據插入數據庫，但問題是

當顯示數據庫中保存的數據時，是否顯示為

• 500 grams Chicken breasts
• 6-8 nos. Green chillies
• 2 tblsp Coriander leaves
• 2 tblsp Mint leaves
• 12 nos. Garlic
• 8-10 nos. Almonds
• 3/4 cup Fresh cream
• 1 inch Ginger
• 1 tblsp Lemon juice
• Oil
• Salt

獲取favirote列表代碼為

-(void)getFavoriteList
{
    if([ArrFav count] > 0)
        [ArrFav removeAllObjects];

    ArrFav = [[NSMutableArray alloc] init];

    sqlite3_stmt *selectStatement=nil;  
    const char *sql ="SELECT Description FROM AddFavoritenew";
    int returnvalue;

    returnvalue = sqlite3_prepare_v2(database, sql, -1, &selectStatement, NULL);

    if(returnvalue==1)
    {
        NSAssert1(0, @"Error: failed to select the database with message '%s'.", sqlite3_errmsg(database));
    }

    if(returnvalue == SQLITE_OK)
    {                       
        while(sqlite3_step(selectStatement) == SQLITE_ROW)
        {   
            char *str = (char *)sqlite3_column_text(selectStatement,0);
            [ArrFav addObject:[NSString stringWithFormat:@"%s",str]];
            NSLog(@"Favorite data is:--> %s",str);
        }       
//        NSLog(@"Favorite data is:--> %@",ArrFav);
    }
}

為什么要從數據庫中獲取數據，例如“，Ä¢”，請注意，我的數據庫描述是文本數據類型.... am也嘗試使用varchar2數據類型

但是我想輸出像保存數據..請幫幫我

感謝和問候

Answer 1

初始字符串為UTF8。

您正在使用此方法從數據庫重新創建它

[ArrFav addObject:[NSString stringWithFormat:@"%s",str]];

這只是用朴素的ascii（ %s格式說明符）轉換str

嘗試使用此方法強制其使用UTF-8解碼字符串：

[ArrFav addObject:[[NSString alloc] initWithCString:str encoding:NSUTF8StringEncoding]];

Answer 2

下面是它的簡單的解決方案，如果你只擔心• symbole。

//Convert • symbole with some different symbole that SQLite DB allowing
- (NSString *) convertToDBFormat:(NSString *)str
{
     NSString *temp = [str stringByReplacingOccuranceOfStrings:@"•" withString:@"!~!"];
    return temp;
}

//Convert back to origional format after retrieving from SQLite DB
- (NSString *) convertBackMyFormat:(NSString *)str 
{
     NSString *temp = [str stringByReplacingOccuranceOfString:@"!~!" withString:@"•"];
    return temp;
}

代替!~! 您可以使用不屬於字符串的任何符號！ :)