找到兩個給定日期

Question

假設之間的天數：

1bhk平200rupee修的

2bhk平400rupee修的費用

假設1個BHK用者自付在某一特定日期的數量。

但是，如果用戶不支付給定日期的金額，然後自動添加20％的額外費用的基本金額。

是否有可能在目標C中實現這一點？

Answer 1

這是非常簡單的兄弟。以下代碼適用於1BHK用戶。

float oneBHKCharge=200; 

float finalDue=0; 

NSString *strDueDate = @"10-02-2017"; 
NSString *strPaidDate = @"10-02-2017"; 

NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init]; 
[dateFormatter setDateFormat:@"dd-MM-yyyy"]; 
NSDate *dueDate = [dateFormatter dateFromString:strDueDate]; 
NSDate *paidDate = [dateFormatter dateFromString:strPaidDate]; 


NSCalendar *gregorianCalendar = [[NSCalendar alloc] initWithCalendarIdentifier:NSCalendarIdentifierGregorian]; 
NSDateComponents *components = [gregorianCalendar components:NSCalendarUnitDay 
                fromDate:dueDate 
                 toDate:paidDate 
                options:0]; 


if (components.day > 0) 
{ 
    finalDue = oneBHKCharge + oneBHKCharge*0.2; 
} 
else 
{ 
    finalDue = oneBHKCharge; 
} 

NSLog(@"%f",finalDue); 

Answer 2

這裏的代碼我解決了這個問題。我在NSUserdefaults中存儲了過期日期。在這裏做支票

- (NSNumber *)invoiceOverDue 
{ 
    NSNumber *result; 
    NSInteger dueDays = [[NSUserDefaults standardUserDefaults] integerForKey:VPPayDue]; 
    if ([[self dueTime] intValue] > dueDays) 
    { 
    result = [NSNumber numberWithBool:YES]; 
    } 

    else { 
    result = [NSNumber numberWithBool:NO]; 
    } 
    return result; 
} 

而這裏的女巫部分我計算逾期時間。

// DueTime calculation 
-(NSNumber *)dueTime 
{ 
// NSTimeInterval secondsPerDay = 24 * 60 * 60; 
if ([[self invoiceStatus] intValue] > 1) 
{ 
    return [NSNumber numberWithInt:0]; 
} 

else { 
    NSTimeInterval calculateTime = [[* Insert here the date of your invoice*] timeIntervalSinceNow]; 
    //DLog(@"dueTime bevor/86400 :%f", calculateTime); 
    calculateTime /= -86400; 
    //DLog(@"dueTime :%f", calculateTime); 
    int returnInt = [ConverterHelper numberRoundUp:calculateTime]; 

    return [NSNumber numberWithInt:returnInt]; 
} 

}