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Top Forums Shell Programming and Scripting Bash variable not being passed Post 302962711 by Aia on Thursday 17th of December 2015 02:53:07 PM
Old 12-17-2015
Instead of this:
Code:
date -d "$id" +%-m-%-d-%Y ; read date #convert to month-date-year

Wouldn't this be what you want?
Code:
date=$(date -d "$id" "+%m-%d-%Y")

This User Gave Thanks to Aia For This Post:
 

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sttime(3)						    ShapeTools Toolkit Library							 sttime(3)

NAME
stMktime, stWriteTime - date and time handling SYNOPSIS
#include <config.h> #include <sttk.h.h> time_tstMktime (char *string); char*stWriteTime (time_t date); DESCRIPTION
stMktime scans the given string and tries to read a date and time from it. It understands various formats of date strings. The following is a list of all valid formats, optional parts in brackets. [Tue] Jan 5[,] [19]93 This includes the standard asctime(3) format. Jan 5 With no year given, the year defaults to the current year. [19]93/01/05 This notation requires month and day represented by exactly two digits. 5.1.[19]93 This is the usual German notation. 5.1. German notation referencing the current year. A certain time, given together with the date must always have the following form. hours:minutes[:seconds] Each of the fields must be an integer value within the proper range (hours: 0-23, minutes and seconds: 0-59). Values below 10 may be written as one digit numbers. The time value may be placed anywhere in the date string: at the beginning, at the end, or somewhere in the middle. Any amount of white- space may be given between a field of the time value and the separating colon. The time is always considered to be local time. stWriteTime generates a time string similar to asctime(3) from its date argument. SEE ALSO
asctime(3) BUGS
Time Zone Names within the time string (like `MET') are not handled properly. In most cases they will cause a failure. sttk-1.7 Thu Jun 24 17:43:35 1993 sttime(3)
All times are GMT -4. The time now is 06:15 PM.
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