shell - 比较文本文件中的日期| shell

Question

我有一个包含第 7 列日期的文件，我的要求是将它与今天的日期进行比较，如果小于它，则删除该完整行。
此外，如果第 7 列中提到的任何日期超过 15 天，则将其修改为最多 15 天
下面的例子-

now="$(date +'%d-%m-%Y')"
now=24-02-2021

文件.txt

abc    xyz        pqr NFS     5       abc.vol
abc    xyz        pqr NFS     50      xyz.bal     23-02-2021
abc    xyz        pqr NFS     5       abcd.xyz
abc    xyz        pqr NFS     15      pqr.vol     25-02-2023
abc    xyz        pqr NFS     5       xyz.bal     24-03-2021
abc    xyz        pqr NFS     3       pqrst.vol   19-01-2019

输出应该是-

abc    xyz        pqr NFS     5       abc.vol
abc    xyz        pqr NFS     5       abcd.xyz
abc    xyz        pqr NFS     15      pqr.vol     11-03-2021
abc    xyz        pqr NFS     5       xyz.bal     24-03-2021

我尝试通过 awk 但它完全不起作用

awk -v d=$now '$7 == "" && $7 < d' file.txt

Answer 1

尝试以下 GNU awk 解决方案:

awk 'NF == 6 { print;next } { OFS="\t";split($7,map,"-");depoch=mktime(map[3]" "map[2]" "map[1]" 00 00 00");if (depoch<systime()) { next };if (+depoch>(systime()+1296000)) { $7=strftime("%d-%m-%Y",(systime()+1296000)) } }1' file.txt

解释:

awk 'NF == 6 { print;                                                    # If there is no date, print the line and skip to the next
               next 
             } 
             { OFS="\t";                                                 # Set the output field delimiter to tab
               split($7,map,"-");                                        # Split the date into the array map (getting day, month and year)
               depoch=mktime(map[3]" "map[2]" "map[1]" 00 00 00");       # Create epoch format of the date
               if (depoch<systime()) { 
                  next                                                   # If epoch format of date is less than today, skip to the next record
               };
               if (depoch>(systime()+1296000)) { 
                  $7=strftime("%d-%m-%Y",(systime()+1296000))            # If the date is greater than today +15 days, set the date to the date + 15 days (1296000 seconds) and format accordingly using strftime
               }  
              }1' file.txt

我假设发布的所需输出不正确。 24-03-2021 比今天晚了 15 天，所以应该改成 11-03-2021