Error with the date command in Bash

Question

I have the below code snippet :

#!/bin/bash
time="date +\"%Y-%m-%d %T,%3N\""
eval echo "`\$time` check"

Actually this is a part of a much larger code . I want to compute the value of time as per the above command on the fly every time the script is run and print the output.

When I run the code this is the below error I get :

date: extra operand ‘%T,%3N"’ Try 'date --help' for more information. check

However when I try to run the command from the command line window it runs fine. The problem is the date has to be in the same format only in the output.

$ date +"%Y-%m-%d %T,%3N"

2017-01-13 11:54:36,604

Please guide me out what should be done in this case.

Thanks in advance for any help

get rid of the '\''s and the extra quotes. time=$(date +"%Y-%m-%d %T,%3N") — David C. Rankin
– David C. Rankin, Commented Jan 13, 2017 at 6:29

David C. Rankin · Accepted Answer · 2017-01-13 06:32:14Z

1

Following from my comment, your problem is that you cannot assign the result of the date command as you have it written, you need:

time=$(date +"%Y-%m-%d %T,%3N")

answered Jan 13, 2017 at 6:32

David C. Rankin

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codeforester · Accepted Answer · 2017-01-13 06:34:57Z

1

Not sure why you need the indirection of eval and backticks. We could simply write:

time=$(date +"%Y-%m-%d %T,%3N")

edited Jan 13, 2017 at 6:34

answered Jan 13, 2017 at 6:31

codeforester

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1 Comment

Subhayan Bhattacharya Over a year ago

Thank you codeforester . I tried your way as well as tried a different way with eval . Both worked :echo "$(date +"%Y-%m-%d %T,%3N") check----" and #!/bin/bash dt="date +\"%Y-%m-%d\"" time="date +\"%T,%3N\"" eval echo "\$dt \$time check"

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