String 查找逗号分隔的字符串元素并删除该行
我有下面的文本文件,希望删除时间(第4个元素)不是15分钟间隔的所有行。下面是一个文件示例:String 查找逗号分隔的字符串元素并删除该行,string,String,我有下面的文本文件,希望删除时间(第4个元素)不是15分钟间隔的所有行。下面是一个文件示例: 22420123252029,1,0,0,0,0.458,8 224,2012,325,2030,1,0,0,0,0.458,9 224,2012,325,2031,1,0,0,0,0.458,9 224,2012,325,2032,1,0,0,0,0.458,9 224,2012,325,2033,1,0,0,0,0.459,8 224,2012,325,2034,1,0,0,0,0.458,8 22
22420123252029,1,0,0,0,0.458,8
224,2012,325,2030,1,0,0,0,0.458,9
224,2012,325,2031,1,0,0,0,0.458,9
224,2012,325,2032,1,0,0,0,0.458,9
224,2012,325,2033,1,0,0,0,0.459,8
224,2012,325,2034,1,0,0,0,0.458,8
224,2012,325,2035,1,0,0,0,0.458,8
224,2012,325,2036,1,0,0,0,0.459,8
224,2012,325,2037,1,0,0,0,0.459,9
224,2012,325,2038,1,0,0,0,0.459,8
224,2012,325,2039,1,0,0,0,0.458,9
224,2012,325,2040,1,0,0,0,0.458,9
224,2012,325,2041,1,0,0,0,0.459,9
224,2012,325,2042,1,0,0,0,0.459,9
224,2012,325,2043,1,0,0,0,0.459,9
224,2012,325,2044,1,0,0,0,0.459,8
224,2012,325,2045,1,0,0,0,0.459,8
224,2012,325,2046,1,0,0,0,0.457,8
删除非15分钟间隔后,更正的文件应为:
224,2012,325,2030,1,0,0,0,0.458,9
224,2012,325,2045,1,0,0,0,0.459,8
既然您还没有为我提供一种语言,我将向您提供用Java实现它的逻辑,以帮助您解决问题。Java不是脚本语言,但不管怎样,它不会有什么坏处:
假设s
是一个值为一行数据的字符串
if (Integer.parseInt(s.split(",")[3].substring(2)) % 15 == 0) {
//You have a row that satisfies your condition
}
如果条件满足,你可以做任何你想做的事。教科书示例:
awk -F, '!$4%15' filename
您希望用什么语言完成此操作?它在Java中相对简单。我假设该值是
HHmm
格式的时间,因此,您不需要进行模数运算,而需要将最后两位数字与15
、30
、45
或00
进行比较。