Question

尝试从字符串（日期）中删除毫秒数

NSString *dateString = @"2016-05-16 13:17:34.674194";
 NSDateFormatter* formater = [[NSDateFormatter alloc] init];
 [formater setDateFormat:@"dd/MM/yyyy HH:mm:ss"];
 NSDate *date = [formater dateFromString:dateString];
 NSLog(@"%@",[formater stringFromDate:date]);

我得到了空

我期待o / p之类的事情（毫秒数）

＆＃34; 06/05/2016 13：17：34＆＃34;

请建议......

Answer 1

复制并粘贴此代码完美运行

NSString *dateString = @" ";
NSDateFormatter* dateFormatter1 = [[NSDateFormatter alloc] init];
dateFormatter1.dateFormat = @"yyyy-MM-dd HH:mm:ss.SSS";
NSDate *yourDate = [dateFormatter1 dateFromString:dateString];
dateFormatter1.dateFormat = @"dd/MM/yyyy HH:mm:ss";
NSLog(@"%@",[dateFormatter1 stringFromDate:yourDate]);

这是输出

2016-05-16 16:43:53.639 StackLearn[6376:151614] 16/05/2016 13:17:34

日期规则格式化程序是您必须设置与字符串相同的日期格式，当您从字符串获取日期时，如果不匹配则您将获得空白

Answer 2

试试这个： -

NSString *dateString = @"2016-05-16 13:17:34.674194";
NSDateFormatter* dateFormatter = [[NSDateFormatter alloc] init];
dateFormatter.dateFormat = @"yyyy-MM-dd HH:mm:ss.SS";     
NSDate *yourDate = [dateFormatter dateFromString:dateString];
dateFormatter.dateFormat = @"dd-MMM-yyyy HH:mm:ss";
NSLog(@"%@",[dateFormatter stringFromDate:yourDate]);

Answer 3

完美的工作

NSString *dateString = @" ";
NSDateFormatter* dateFormatter1 = [[NSDateFormatter alloc] init];
dateFormatter1.dateFormat = @"yyyy-MM-dd HH:mm:ss.SSS";
NSDate *yourDate = [dateFormatter1 dateFromString:dateString];
dateFormatter1.dateFormat = @"dd/MM/yyyy HH:mm:ss";
NSLog(@"%@",[dateFormatter1 stringFromDate:yourDate]);

从iOS中的String中删除milliSeconds

3 个答案: