ios 在目标 C 中转换日期格式
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Converting a date format in objective C
提问by farajnew
How can i convert the following date "Sun Jul 17 07:48:34 +0000 2011" to the following format "2011-07-17 07:48:34"?
如何将以下日期“Sun Jul 17 07:48:34 +0000 2011”转换为以下格式“2011-07-17 07:48:34”？
I used NSDateFormatteras shown below but it didnt work. It gives null as a result.
我使用NSDateFormatter如下所示，但没有用。结果是 null 。
NSDateFormatter *objDateFormatter = [[NSDateFormatter alloc] init];
[objDateFormatter setDateFormat:@"yyyy-MM-dd HH:mm:ss"];
[objDateFormatter dateFromString:sDate];
回答by Suhail Patel
You've got your Date Format wrong for the style of Date you are passing in. Here is a document explaining the different modifiers: Date Format Patterns
对于传入的日期样式，您的日期格式错误。这是解释不同修饰符的文档：日期格式模式
To parse the Date "Sun Jul 17 07:48:34 +0000 2011", you'd need a Format like so:
要解析日期“Sun Jul 17 07:48:34 +0000 2011”，您需要如下格式：
[dateFormat setDateFormat:@"EEE MMM dd HH:mm:ss ZZZ yyyy"];
To get it into the following format: "2011-07-17 07:48:34", here is the full code:
要将其转换为以下格式：“2011-07-17 07:48:34”，这里是完整代码：
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:@"EEE MMM dd HH:mm:ss ZZZ yyyy"];
NSDate *date = [dateFormatter dateFromString:sDate];
// Convert to new Date Format
[dateFormatter setDateFormat:@"yyyy-MM-dd HH:mm:ss"];
NSString *newDate = [dateFormatter stringFromDate:date];
回答by steipete
If sDate is your string that's in the format "Sun Jul 17 07:48:34 +0000 2011", you have to convert that into a NSDate. Then, you can use a second NSDateFormatter to convert this date into your desired format "yyyy-MM-dd HH:mm:ss".
如果 sDate 是格式为“Sun Jul 17 07:48:34 +0000 2011”的字符串，则必须将其转换为 NSDate。然后，您可以使用第二个 NSDateFormatter 将此日期转换为您想要的格式“yyyy-MM-dd HH:mm:ss”。
If you need to figure out the string needed to convert, this site has a great reference: http://unicode.org/reports/tr35/tr35-6.html#Date_Format_Patterns
如果您需要弄清楚需要转换的字符串，这个站点有一个很好的参考：http: //unicode.org/reports/tr35/tr35-6.html#Date_Format_Patterns
e.g. "Sun Jul 17 07:48:34 +0000 2011" can be parsed with
例如“Sun Jul 17 07:48:34 +0000 2011”可以解析为
[dateFormatter setDateFormat:@"EEE MMM d HH:mm:ss ZZZ yyyy"];
As a bit of advice, non-standard dates like "Jul" (should be July) can make problems here. It's faster to just convert the needed parts in plain C, with strftime(). See the reference: http://publib.boulder.ibm.com/infocenter/iadthelp/v7r0/index.jsp?topic=/com.ibm.etools.iseries.langref.doc/rzan5mst263.htm
作为一点建议，像“七月”（应该是七月）这样的非标准日期可能会在这里出现问题。使用 strftime() 在普通 C 中转换所需的部分会更快。请参阅参考资料：http: //publib.boulder.ibm.com/infocenter/iadthelp/v7r0/index.jsp?topic=/com.ibm.etools.iseries.langref.doc/rzan5mst263.htm
You can then convert the unix timestamp back in a NSDate with
然后，您可以将 unix 时间戳转换回 NSDate 中
[NSDate dateWithTimeIntervalSince1970:]
Using C for date/time parsing is usually up to 10x faster. NSDateFormatter is a slow dog.
使用 C 进行日期/时间解析通常快 10 倍。NSDateFormatter 是一条慢狗。
