CODEWITHSUNDEEP
shell
modusTollens
modusTollens

Reputation: 417

Shell script: get date as variable

I am trying to create a quick script that will get a file from a backup folder. The backup file is generated every day and have the filename format of:

backup-sql-YYYYMMDD.zip

I am trying to get the filename by using:

#!/bin/sh

VAR_1="backup-sql-"
VAR_2=date +%Y%m%d
VAR_3=$VAR_1$VAR_2

echo "$VAR_3"

However the output is:

backup-sql-

When I run:

date +%Y%m%d

I get

20180704

So I am not sure why this is happening. Please can someone help me :)

Upvotes: 0

Views: 7677

Answers (2)

nbari
nbari

Reputation: 26925

You could use:

$(command)

or

`command`

From your example give a try to:

VAR_2=$(date +%Y%m%d)

Check Command Substitution for more details:

When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by ‘$’, ‘`’, or ‘\’. The first backquote not preceded by a backslash terminates the command substitution. When using the $(command) form, all characters between the parentheses make up the command; none are treated specially.

Upvotes: 1

asatsi
asatsi

Reputation: 472

You should be using backticks to capture the output of the date command and assign it to a shell variable.

VAR2=`date "+%Y%m%d"`

In case you want to make sure the variable value is passed on to subsequent child processes you may want to export it as:

export VAR2=`date "+%Y%m%d"`

Upvotes: 1

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