Bash – Unix epoch in milliseconds

bashdateshell-scripttimetimestamps

I'm prepending the Unix epoch with "nanosecond" precision into output of my command as below:

$ command | while read line; do d=`date +%s%N`; echo $d $line; done > file

I looked around to find out how to turn "nanosecond" into "millisecond". For example, I followed the solution given here. So, I tried both suggested approaches:

$ command | while read line; do d=`echo $(($(date +%s%N)/1000000))`; echo $d $line; done > file
$ command | while read line; do d=`date +%s%N | cut -b1-13`; echo $d $line; done > file

However, in both cases when I insert the file into InfluxDB and query my database I get this:

time    
1970-01-01T00:24:05.419982325Z  
1970-01-01T00:24:05.419982344Z  
1970-01-01T00:24:05.419982371Z  
1970-01-01T00:24:05.419982378Z  
1970-01-01T00:24:05.419982388Z  
1970-01-01T00:24:05.419982401Z

Update

When I use epoch with nanosecond accuracy date +%s%N, I get this:

time
2015-10-21T08:59:59.978902683Z  
2015-10-21T08:59:59.982615836Z  
2015-10-21T08:59:59.983958069Z  
2015-10-21T08:59:59.98805317Z   
2015-10-21T08:59:59.99717678Z   
2015-10-21T09:00:00.028624495Z

I'm expecting such an output:

2015-10-21T09:12:10.001327Z

Please let me know if you have any solution.

Best Answer

You could use bc and printf:

printf "%0.f" "$(bc <<<"$(date +"%s.%N")*1000")"

This gives the number of miliseconds since january 1970. I didn't use the scale=n option of bc on purpose, because that would not round the value, instead it cuts the rest away (I know, it's pedantic). bc reads from file or from the standard input. <<< is a here string which expands the contents and supplies them as standard input to bc. This is given to printf to round the value.

See this as an example:

$ d=$(date "+%s.%N")
$ echo $d; bc <<<"scale=0; ($d*1000)/1"; printf "%0.f" "$(bc <<<"$d*1000")"
1445423229.512731661 # plain date
1445423229512 # bc with scale
1445423229513 # bc/printf

In the loop it would then look like this:

command | while read line; do
  d=$(printf "%0.f" "$(bc <<<"$(date +"%s.%N")*1000")")
  echo "$d $line"
done >file

Related Solutions

Bash – How to Name a File in the Deepest Level of a Directory Tree

That's an odd request!

I'd use find + awk to grab a file in the deepest directory:

bash-3.2$ deepest=$(find / -type f | awk -F'/' 'NF > depth {
>     depth = NF;
>     deepest = $0;
> }
>
> END {
>     print deepest;
> }')

Using ${deepest} in your mv command is left as an exercise but the following five lines may help you further:

bash-3.2$ echo "${deepest}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina/Buenos_Aires.rb

bash-3.2$ echo "${deepest%.*}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina/Buenos_Aires

bash-3.2$ echo "${deepest%/*}"
/Developer/SDKs/MacOSX10.6.sdk/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib/ruby/gems/1.8/gems/activesupport-2.3.5/lib/active_support/vendor/tzinfo-0.3.12/tzinfo/definitions/America/Argentina

bash-3.2$ echo "${deepest##*/}"
Buenos_Aires.rb

bash-3.2$ echo "${deepest##*.}"
rb

Following update to question:

find -type d [...] "This would only find the directory. [...] How could this be solved in the most simple way?".

By supplying -type f to find to find all files (f), not all directories (d).

Bash – How to Get Milliseconds Since Unix Epoch

date +%s.%N will give you, eg., 1364391019.877418748. The %N is the number of nanoseconds elapsed in the current second. Notice it is 9 digits, and by default date will pad this with zeros if it is less than 100000000. This is actually a problem if we want to do math with the number, because bash treats numbers with a leading zero as octal. This padding can be disabled by using a hyphen in the field spec, so:

echo $((`date +%s`*1000+`date +%-N`/1000000))

would naively give you milliseconds since the epoch.

However, as Stephane Chazelas points out in comment below, that's two different date calls which will yield two slightly different times. If the second has rolled over in between them, the calculation will be an entire second off. So:

echo $(($(date +'%s * 1000 + %-N / 1000000')))

Or optimized (thanks to comments below, though this should have been obvious):

echo $(( $(date '+%s%N') / 1000000));

Best Answer

Related Solutions

Bash – How to Name a File in the Deepest Level of a Directory Tree

Following update to question:

Bash – How to Get Milliseconds Since Unix Epoch

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