Linux Find a date for a day file __linux

  find a file with a date of the day

Skill Manager (2000-12-22-23:24)

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A= ' Find ~-print ' | Ls-l--full-time $A 2>/dev/null | grep "June 27" | grep 1998 Linux commands are very powerful. Write the following script for you, with a little more convenience. You can save it as a file of any name, and put the X attribute on it. #!/bin/sh # The right of usage, distribution and modification are here with granted by the author. # The author deny any responsibilities and liabilities related to the code. # Ok=0 A= ' Find $1-print ' If expr $ = 1 >/dev/null; Then M=jan; ok=1; Fi If expr $ = 2 >/dev/null; Then M=feb; ok=1; Fi If expr $ = 3 >/dev/null; Then M=mar; ok=1; Fi If expr $ = 4 >/dev/null; Then M=APR; ok=1; Fi If expr $ = 5 >/dev/null; Then M=may; ok=1; Fi If expr $ = 6 >/dev/null; Then M=jun; ok=1; Fi If expr $ = 7 >/dev/null; Then M=jul; ok=1; Fi If expr $ = 8 >/dev/null; Then M=aug; ok=1; Fi If expr $ = 9 >/dev/null; Then M=sep; ok=1; Fi If expr $ = >/dev/null; Then m=oct; ok=1; Fi If expr $ = >/dev/null; Then M=nov; ok=1; Fi If expr $ = >/dev/null; Then M=dec; ok=1; Fi If expr $ = 1 >/dev/null; Then M=jan; ok=1; Fi If expr $OK = = 1 >/dev/null; Then Ls-l--full-time $A 2>/dev/null | grep "$M $" | grep $; Else echo Usage: $ path year Month day; echo Example: $ ~ 1998 6 30;         fi