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  1. #!/bin/bash
  2.  
  3. uname -a
  4.  
  5. echo "*** date in a loop"
  6.  
  7. base=$(date +"%s")
  8. for((i=0; i<5; ++i)); do
  9.     date -d @$((i * 24 * 60 * 60 + base)) +%Y-%m-%d
  10. done
  11.  
  12. echo "*** perl"
  13.  
  14. perl -le 'use POSIX qw(strftime);
  15.     $t = time;
  16.     for $_ (1..5) {
  17.         print strftime("%Y-%m-%d", localtime($t));
  18.         $t += 24 * 60 * 60 }'
  19.  
Success #stdin #stdout 0.02s 6868KB
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stdin
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stdout
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Linux checker 4.19.0-14-amd64 #1 SMP Debian 4.19.171-2 (2021-01-30) x86_64 x86_64 x86_64 GNU/Linux
*** date in a loop
2022-05-08
2022-05-09
2022-05-10
2022-05-11
2022-05-12
*** perl
2022-05-08
2022-05-09
2022-05-10
2022-05-11
2022-05-12
https://ideone.com/WkAoib
https://stackoverflow.com/questions/72155088/how-can-i-get-a-sequence-of-dates-in-bash
language:
Bash (bash 5.0.3)
created:
3 years ago
visibility:
public

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