Passing timestamp variable from shell to gawk

Question

I want to take time difference of two time stamp but getting an error while passing timestamp variable from shell toawk.

Shell code:

FTIMESTAMP="2015-07-01 12:30:50"
gawk -v FTIMESTAMP=$FTIMESTAMP -v DSECONDS=$DSECONDS -f test.awk /home/abc/TShift.csv

Error is:

gawk: 12:30:50
gawk:   ^ syntax error

Escaping character is also not working FTIMESTAMP="2015-07-01 12\:30\:50".

I have another timestamp in awk and want to take time difference between them.

See how that's one of the examples at [Security implications of forgetting to quote a variable in bash/POSIX shells](http://unix.stackexchange.com/q/171346) — Stéphane Chazelas, Oct 09 '15 at 06:58

score 2 · Accepted Answer · edited Apr 13 '17 at 12:36

2

You need to quote variables to prevent shell from performing split+glob:

gawk -v FTIMESTAMP="$FTIMESTAMP" -v DSECONDS="$DSECONDS" ...

A note that -v var="$shell_var" will expand escape sequences in $shell_var. You need to use ENVIRON or ARGV variables to pass $shell_var as-is from shell to awk.

edited Apr 13 '17 at 12:36

Community

1

answered Oct 09 '15 at 06:59

cuonglm

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score 0 · Answer 2 · answered Oct 09 '15 at 06:59

try either

gawk -v FTIMESTAMP="$FTIMESTAMP" -v DSECONDS=$DSECONDS -f test.awk /home/abc/TShift.csv

there is a white in $FTIMESTAMP

or in calling shell

FTIMESTAMP="2015-07-01 12:30:50"
export FTIMESTAMP

and in awk's script (in BEGIN { .. } section)

FTIMESTAMP=ENVIRON["FTIMESTAMP"] ;

Passing timestamp variable from shell to gawk

2 Answers2