I want to select a file name with a date that I generated in the past in the script.

Currently, there is a file that was granted using the date command.
■ Example: 20180101.zip to 20180131.zip

From 9:00 on the selected date to 9:00 on the next day
without unzipping the zip, the line that corresponds to a particular string is
I wrote a script to find out how many lines there are, but it doesn't work.
Could you teach me?

When you want to display consecutive integers in the script, please let me know if there is a way to represent 01-31 instead of 1-31.
In order to select the two files to be referenced, I would like to see a configuration that can output a difference such as aa+1.

■ Command Example:

#!/bin/sh
i = 10
j = `$i+1`
while [$i-ne20];
do
echo"`ls201801"$i".zip201801"$j".zip | xargs-n1-IX unzip-p X | awk-F-"Dec"$i"09:00:00"<=$1&&$j"09:00"'|grep', allow,'|wc-l`"
  i = `expr$i+1`
  j = `expr$j+1`
done

Thank you for your cooperation.

linux shellscript

2022-09-30 21:30

2 Answers

When you want to display consecutive integers in the script, please let me know if there is a way to represent 01-31 instead of 1-31.

You can generate consecutive numbers with the seq command.I was able to format my version with -f.

$seq-f"%02.f"110
01
02
03
04
05
06
07
08
09
10

If the file name is the date, you can just pass it over.

$seq2018011020180120
20180110
20180111
20180112
20180113
20180114
20180115
20180116
20180117
20180118
20180120

In order to select the two files to be referenced, I would like to see a configuration that can output a difference such as aa+1.

You can use the expr command to calculate expressions expressed in strings.

$expr "1+1"
2

Combine for, seq, and expr

$for i in `seq2018011020180120`; do echo$i`expr$i+1`; done
20180110 20180111
20180111 20180112
20180112 20180113
20180113 20180114
20180114 20180115
20180115 20180116
20180116 20180117
20180117 20180118
20180118 20180119
20180119 20180120
20180120 20180121

2022-09-30 21:30

How many lines of content correspond to a particular string without unzipping the zip from 9:00 on the selected date to 9:00 on the next day

Instead of selecting a date, I thought about aggregating it by date (from the beginning to the end of the month).

unzip-p'201801*.zip'|
awk-F'[:]'
    $2>1&&$3<9&&$0~", allow, "{cnt[$2-1]++}
    $3>=9&&$0 ~", allow, "{cnt[$2+0]++}
  END{
    for(dincnt){
      printf("201801%02d:%d\n", d, cnt[d])
    }
  }
'

The unzip command argument is enclosed in a single quote ('201801*.zip') because it does not deploy wildcards (*) to the shell (expanded within unzip).
The awk script aggregates the number from midnight to 9:00 a.m. (cnt[$2-1]++) for the previous day ($2>1 to skip from midnight to 9:00 a.m. on January 1st (early Monday).(The code in the questionnaire says Dec.)…).

2022-09-30 21:30

If you have any answers or tips

Popular Tags

python x 4647

android x 1593

java x 1494

javascript x 1427

c x 927

c++ x 878

ruby-on-rails x 696

php x 692

python3 x 685

html x 656