CODEWITHSUNDEEP
bash
user3053231
user3053231

Reputation:

Printing day of the week from exact date in bash

I am trying to print the day of the week of given date. This command works pretty well:

TARGET=$(date -u -d'2015-10-25' '+%u')

But inside my bash script there is an error, what should be wrong? 

#!/bin/bash
day=25
month=10
year=2015
command1='date -u -d'
command3=''\'
command2=$year-$month-$day
fullcommand=$command1$command3$command2$command3' '$command3'+%u'$command3
echo $fullcommand
TARGET=$($fullcommand)
echo $TARGET

There is an error: 

date: the argument ‘'+%u'’ lacks a leading '+';

Upvotes: 4

Views: 112

Answers (2)

anubhava
anubhava

Reputation: 785098

No need to use so many temporary variables and definitely escaping single quote inside another single quote won't work in shell.

Simplify it like this:

#!/bin/bash
day=25
month=10
year=2015
command1='date -u -d'
TARGET=$(date -u -d "$year-$month-$day" '+%u')
echo $TARGET

Upvotes: 2

Hasse Björk
Hasse Björk

Reputation: 1601

This works:

#!/bin/bash
day=25
month=10
year=2015
command1='date -u -d'
command3=''\'
command2=$year-$month-$day
fullcommand="$command1 $command2 +%u"
echo $fullcommand
TARGET=$($fullcommand)
echo $TARGET

I do not have the answer to why, though

Upvotes: 1

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