Question

我有一个时间戳如下。它采用字符串格式 “/日期（1402987019190 + 0000）/” 我想将其转换为NSDate。有人可以帮助我进入这个吗？

Answer 1

使用dateWithTimeIntervalSince1970:，

NSDate *date = [NSDate dateWithTimeIntervalSince1970:timeStamp];
NSLog(@"%@", date);

使用此函数将字符串转换为时间戳Ref。 https://stackoverflow.com/a/932130/1868660

-(NSString *)dateDiff:(NSString *)origDate {
    NSDateFormatter *df = [[NSDateFormatter alloc] init];
    [df setFormatterBehavior:NSDateFormatterBehavior10_4];
    [df setDateFormat:@"EEE, dd MMM yy HH:mm:ss VVVV"];
    NSDate *convertedDate = [df dateFromString:origDate];
    [df release];
    NSDate *todayDate = [NSDate date];
    double ti = [convertedDate timeIntervalSinceDate:todayDate];
    ti = ti * -1;
    if(ti < 1) {
        return @"never";
    } else  if (ti < 60) {
        return @"less than a minute ago";
    } else if (ti < 3600) {
        int diff = round(ti / 60);
        return [NSString stringWithFormat:@"%d minutes ago", diff];
    } else if (ti < 86400) {
        int diff = round(ti / 60 / 60);
        return[NSString stringWithFormat:@"%d hours ago", diff];
    } else if (ti < 2629743) {
        int diff = round(ti / 60 / 60 / 24);
        return[NSString stringWithFormat:@"%d days ago", diff];
    } else {
        return @"never";
    }   
}

Answer 2

如果我正确理解了您的意思，您需要使用给定格式解析字符串中的时间戳。您可以使用正则表达式来提取时间戳值。

NSString * timestampString = @"/Date(1402987019190+0000)/";
NSString *pattern = @"/Date\\(([0-9]{1,})\\+0000\\)/";

NSRegularExpression *regex = [NSRegularExpression
 regularExpressionWithPattern:pattern
                      options:NSRegularExpressionCaseInsensitive
                        error:nil];
NSTextCheckingResult *textCheckingResult = [regex firstMatchInString:timestampString options:0 range:NSMakeRange(0, timestampString.length)];
NSRange matchRange = [textCheckingResult rangeAtIndex:1];
NSString *match = [timestampString substringWithRange:matchRange];

NSDate *date = [NSDate dateWithTimeIntervalSince1970:[match intValue];
NSLog(@"%@", date);

如何将时间戳转换为nsdate

2 个答案: