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我有一个字符串“2012-09-16 23:59:59 JST”我想将此日期字符串转换为 NSDate。

NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"yyyy-MM-dd hh:mm:ss Z"];
NSDate *capturedStartDate = [dateFormatter dateFromString: @"2012-09-16 23:59:59 JST"];
NSLog(@"%@", capturedStartDate);

但它不起作用。它给出空值。请帮忙..

4

3 回答 3

37

使用 24 小时制时,小时说明符必须是大写 H,如下所示:

[dateFormatter setDateFormat:@"yyyy-MM-dd HH:mm:ss Z"];

在此处检查正确的说明符:http ://unicode.org/reports/tr35/tr35-6.html#Date_Format_Patterns

但是,您需要为日期格式化程序设置语言环境:

// Set the locale as needed in the formatter (this example uses Japanese)
[dateFormat setLocale:[[NSLocale alloc] initWithLocaleIdentifier:@"ja_JP"]];

完整的工作代码:

NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"yyyy-MM-dd HH:mm:ss zzz"];
[dateFormatter setLocale:[[NSLocale alloc] initWithLocaleIdentifier:@"ja_JP"]];
NSDate *capturedStartDate = [dateFormatter dateFromString: @"2012-09-16 23:59:59 JST"];
NSLog(@"Captured Date %@", [capturedStartDate description]);

输出(格林威治标准时间):

Captured Date 2012-09-16 14:59:59 +0000
于 2012-09-14T06:34:04.960 回答
6
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"yyyy-MM-dd HH:mm:ss Z"];
NSDate *capturedStartDate = [dateFormatter dateFromString: @"2012-09-16 23:59:59 GMT-08:00"];
NSLog(@"%@", capturedStartDate);
于 2012-09-14T06:50:34.413 回答
1
NSString *dateString = @"01-02-2010";
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];

// this is imporant - we set our input date format to match our input string
// if format doesn't match you'll get nil from your string, so be careful

[dateFormatter setDateFormat:@"dd-MM-yyyy"];
NSDate *dateFromString = [[NSDate alloc] init];
 // end
dateFromString = [dateFormatter dateFromString:dateString];
[dateFormatter release];
于 2012-09-14T06:47:13.510 回答