我试图用以下格式将字符串解析成日期时间:
[日],[日期][月][年][时间][上午/下午][时区](例如:)
"2011年12月1日,星期四,上午8: 30,下午8:
我用“ DateTime” 来做这个。 使用格式分析( “ dddd, dd MM yyyyy hh:mmtt” ) 格式。 但是, Timmzone 给了我一个问题。 没有用于读取以这种方式写成的时区代码 。 我不管时间区, 所以我想把它删掉或者读读它 - 只要异性性行为有效。
去除它的一种方法是将它从字符串中(使用.replace)实际删除(使用.replace)——然而我不知道源代码将产生多少个不同的时区,反正我认为长长的替换线看起来丑陋且容易出错。
难道有办法去掉它,或阅读它吗?
您可以扫描字符串中的空格, 并在第五个空格之后切除所有空格。 如果只有四个空格, 请保留整个字符串( 这意味着没有时区 ) 。
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