problem with shell variables in awk

Question

I have a command which is executed successfully on command line:

ls -l | awk '/Sep 26/{split($8,a,":");if(a[1]a[2]>=1045 && a[1]a[2]<=1145)print $9}'

I am including the same thing in a shell script below:

#!/bin/ksh

date1=$1
date2=$2
time1=$3
time2=$4
ls -l| awk -v d1=${date1} -v d2=${date2}  -v t1=${time1} -v t2=${time2} '/d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 && a[1]a[2]<=t2) print $9}'

But this does not work.please see below the execution

ksh -vx test.sh Sep 26 1045 1145
#!/bin/ksh

date1=$1
+ date1=Sep
date2=$2
+ date2=26
time1=$3
+ time1=1045
time2=$4
+ time2=1145
ls -l| awk -v d1=${date1} -v d2=${date2}  -v t1=${time1} -v t2=${time2} '/d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 &&a[1]a[2]<=t2)print $9}'
+ awk -v d1=Sep -v d2=26 -v t1=1045 -v t2=1145 /d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 &&a[1]a[2]<=t2)print $9}
+ ls -l
awk: syntax error near line 1
awk: bailing out near line 1

I am using Solaris OS:I have tried with nawk now but there is no error but also there is no output.

pearl[ncm_o11.2_int.@].293> ksh -vx test.sh Sep 26 1045 1145
#!/bin/ksh

date1=$1
+ date1=Sep
date2=$2
+ date2=26
time1=$3
+ time1=1045
time2=$4
+ time2=1145
ls -l| nawk -v d1=${date1} -v d2=${date2}  -v t1=${time1} -v t2=${time2} '/d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 &&a[1]a[2]<=t2)print $9}'
+ ls -l
+ nawk -v d1=Sep -v d2=26 -v t1=1045 -v t2=1145 /d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 &&a[1]a[2]<=t2)print $9}

When i am executing with out the shell variables inside the script.its executing perfectly.

*Note:*I am using Solaris OS.

I figured out the problem lies in the final framed command inside shell script:

ls -l|nawk -v d1=Sep -v d2=26 -v t1=1045 -v t2=1145 '/d1 d2/{split($8,a,":");if(a[1]a[2] >=t1 && a[1]a[2]<=t2)print $9}'

But i am not sure why this is failing to give the correct output with -v flags

awk --v. also, does this anything to do with shell variables? does it work if you call it directly without variable substitution? — Karoly Horvath
– Karoly Horvath, Commented Sep 26, 2011 at 8:51

Dimitre Radoulov · Accepted Answer · 2011-09-26 13:37:10Z

2

As already suggested, if you need a dynamic regular expression, you need to use $0 ~ d1 " " d2 instead of /d1 d2/.

answered Sep 26, 2011 at 13:37

Dimitre Radoulov

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Dimitre Radoulov · Accepted Answer · 2011-09-26 08:47:56Z

0

Which operating system? It's possible that when you run the command interactively awk points to a different awk implementation. If you're on Solaris, for example, try running your script with nawk (or gawk), instead of awk.

answered Sep 26, 2011 at 8:47

Dimitre Radoulov

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6 Comments

Dimitre Radoulov Over a year ago

So try using nawk in your script.

Dimitre Radoulov Over a year ago

At least the error is gone :) I suppose that /d1 d2/ doesn't mean what you expect. If you need a dynamic regular expression, you need to use something like this: $0 ~ d1 " " d2.

Vijay Over a year ago

I have observed that the command is not returning any out put.:ls -l|nawk -v d1=Sep -v d2=26 -v t1=1045 -v t2=1145 '/d1 d2/{split($8,a,":");if(a[1]a[2]>=t1 && a[1]a[2]<=t2)print $9}'

Dimitre Radoulov Over a year ago

Did you try to change /d1 d2/ to $0 ~ d1 " " d2?

glenn jackman Over a year ago

@Dimitre, you have the right answer with $0 ~ d1 " " d2: post it as an answer to receive proper credit.

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