awk script: need help


 
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# 8  
Old 07-15-2016
Try
Code:
awk '
                        {OUT[$1] = $2
                        }
/eof/ && 
OUT["time"] <= 12       {delete OUT["eof"]
                         DL = ""
                         for (o in OUT) {printf "%s%s=%s", DL, o,  OUT[o]
                                         DL = ","
                                        }
                         printf RS
                        }
' FS="=" OFS="," file
name=admin,transid=01,message=test,time=06.58.51
name=bhu3,transid=03,message=testing,time=07.58.51


Last edited by RudiC; 07-15-2016 at 09:24 AM..
# 9  
Old 07-15-2016
Hi RudiC,

I tried your code but the output is not as expected.

Code:
[bhupesh@RHL9 capgemini]$ awk '
                        {OUT[$1] = $2
                        }
/eof/ &&
OUT["time"] <= 12       {delete OUT["eof"]
                         DL = ""
                         for (o in OUT) {printf "%s%s=%s", DL, o,  OUT[o]
                                         DL = ","
                                        }
                         printf RS
                        }
' FS="=" OFS="," log
name=admin,id=01,time=06.58.51,message=test
transid=03,name=bhu3,id=01,time=07.58.51,message=testing

Please suggest on the same.

Thanks.

Regards,
Bhupesh
# 10  
Old 07-15-2016
Hi Bhupesh,
For the comparison, can you please check whether the following does meet your need (untested)
Code:
awk '
BEGIN {FS = "\n"
OFS = ","
RS = ""
}
 {trn=$1;nm=$2;tme=$3;msg=$4;
split($3,a,"=");
split(a[2],b,".");
if ((b[1] + 0) >=0 && (b[1] + 0) <= 12)
{ 
print $1,$2,$3,$4
}}' log3

This User Gave Thanks to chill3chee For This Post:
# 11  
Old 07-15-2016
Hi Chill3chee,

It worked, the output is as expected.

The only doubt is : if ((b[1] + 0) >=0 && (b[1] + 0) <= 12) , i presume you are doing this :b[1] + 0 just to ensure that value is numeric by adding it to a 0 any string will also become 0.Please correct if i am wrong.

I tried this and it worked :
Code:
awk '
BEGIN {FS = "\n"
OFS = ","
RS = "eof\n"
}
 {trn=$1;nm=$2;tme=$3;msg=$4;
split($3,a,"=");
split(a[2],b,".");
if (b[1] <= 12)
{ 
print $1,$2,$3,$4
}}' log

output :
Code:
transid=01,name=admin,time=06.58.51,message=test
transid=03,name=bhu3,time=07.58.51,message=testing

so now i dont have to replace eof with empty string. eof\n worked for me..Smilie Smilie

Thank you very much Don,RudiC and Chill3chee for your help.

Cheers !!!!

Regards,
Bhupesh
# 12  
Old 07-15-2016
That could be simplified a little bit to:
Code:
awk '
BEGIN {	FS = "\n"
	OFS = ","
	RS = "eof\n"
}
{	split($3, a, /[=.]/)
	if((a[2] + 0) <= 12)
		print $1, $2, $3, $4
}' log

The reason chill3chee added 0 before doing the comparison is because the default type for a field created as the result of a split is string; not number. If you were looking for times before 9am, using:
Code:
	if(a[2] <= 9)

you'd get the wrong results sometimes because in a string comparison, "10" is less than "9". But with:
Code:
	if((a[2] + 0) <= 9)

you force a numeric comparison and get the results you want.
This User Gave Thanks to Don Cragun For This Post:
# 13  
Old 07-15-2016
Try
Code:
awk '
/eof/   {if (HR <= 12)   print TMP
         TMP = DL = ""
         next
        }
        {TMP = TMP DL $0
         DL  = ","
        }
/^time/ {HR = substr ($0, 6, 2)
        }
' file
transid=01,name=admin,time=06.58.51,message=test
transid=03,name=bhu3,time=07.58.51,message=testing

# 14  
Old 07-15-2016
Thank you Don.. Hi RudiC, Your code is working but i am not able to understand the code except for the next statement which tells awk to read the next line and substr line in which you are generating the value for HR . The comparison is done earlier and the HR variable is initialized later, it is confusing for me, please explain the code. Thanks
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