Error when executing script

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Operating Systems Linux Red Hat Error when executing script
# 1  
Error when executing script


I wrote this script to test if the output for DIR1 and DIR2 comes out as I want :


DAY=$(date +%d)
MONTH=$(date +%b)
YEAR=$(date +%Y)

echo $DIR1
echo $DIR2

This is the output I want for echo $DIR1 : Apr2017_Blast_BC01
And this is the output I want for echo $DIR2 : Apr2017_Blast_BC15

This is the error I get when I execute the script :

[root@L28tstream1 ~]# ./
./ line 6: Apr2017_Blast_BC01: command not found
./ line 7: Apr2017_Blast_BC15: command not found

What is wrong with the syntax in the script above?
# 2  

the construct $(command) is called command substitution and evaluates to the output of command and you use it correctly for the assignment of the variables DAY, MONTH and YEAR. You do not want to use command substitution when you assign the values for DIR1 and DIR2.

DAY=$(date +%d)
MONTH=$(date +%b)
YEAR=$(date +%Y)

echo $DIR1
echo $DIR2

# 3  
Thanks! That did the trick!
# 4  
It would be safer to wrap the entire assignment in double quotes as a common practice so that if you accidentally assign a space to either variable, then the assignment to DIR1 or DIR2 will not fail:-
LO_VAR="Text and space"

MY_VAR2=${HI_VAR}${LO_VAR}"_other_text"       # May fail in some shells

I always clearly mark the variable name with curly braces { & } to ensure I don't confuse the variable with any literal text or underscores that follow. You can get trapped with things like this:-
/home/rbatte1> V1="Hello"
/home/rbatte1> V2="$V1_Goodbye"
/home/rbatte1> echo "\$V2=\"$V2\""

This is because the assignment is looking for variable V1_Goodbye (equivalent to ${V1_Goodbye} ) which has not been set.

I would also suggest that you could consolidate commands such that you just do this:-
MONTHYEAR=$(date +%b%Y)

Of course, it depends what you use the values for elsewhere.

I hope that this helps,
# 5  
Yes, a consolidation makes sense.
MONTHYEAR=$(date +%b%Y)

""Quoting of variables is never a mistake. Especially beginners should do it always!
But - now for the advanced user - the quoting of an assignment value is only needed if there is a literal space.
DIR3="${MONTHYEAR} Blast BC15"
echo "$DIR3"

Not if the space comes from a variable
NAME="Blast BC15 *"
echo "$DIR3"

The latter is not an assignment - the argument for the echo command must be quoted!
case ${MONTHYEAR}_$NAME in ...

[[ ${MONTHYEAR}_$NAME ... ]]

In all other cases the shell does first variable substitution then word splitting then globbing.
Last but not least, the [ ] is a false friend:
[ "${MONTHYEAR}_$NAME" ... ]

In fact the [ is a command and its arguments must be quoted!
# 6  
I have another question on command substitution.
The same script, but different command substitution.

I simply run this command from the terminal :

[root@L28tstream1 emokheng]# stat -c %z * | awk '{print $1}'
[root@L28tstream1 emokheng]#

As you see, I get the dates of the file generated in the directory.

However, when I attempt to use this command as a substitute, like this, I get a diferent output :

[root@L28tstream1 emokheng]# D=$(stat -c %z * | awk '{print $1}')
[root@L28tstream1 emokheng]# echo $D
2015-08-18 2015-08-18

It all displays in one line. Why is this, and how do I get it to display in two separate lines like the command above this?

Actually the main output I want is to get ONLY the day section of the whole date. Means I want to extract only '18' from 2015-08-18. This is for all the files in the directory.
Therefore, I use this command :

[root@L28tstream1 emokheng]# FDAY=$(date -d "$D" '+%d')
date: invalid date `2015-08-18\n2015-08-18'
[root@L28tstream1 emokheng]#

However, as you can see, the error is shown. Perhaps if
echo $D

displays two separate lines, the error above will not happen?

How do I display ONLY the day part of the whole date for a file in a directory?

Last edited by anaigini45; 05-05-2017 at 07:12 AM..
# 7  
The -d option of date takes one single date only, not several separated by a line feed char. How about
D="$(stat -c %z sh* | awk -F"[ -]" '{print $3}')"
echo "$D"

This User Gave Thanks to RudiC For This Post:
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