Count delimiter(~|*) each row in a file and return 1 or 0


 
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# 1  
Old 10-23-2015
Count delimiter(~|*) each row in a file and return 1 or 0

Hi

I want to check delimiter in file. Delimiter in my file is ~|*

sample of file :
Code:
ABC~|*edgf~|*T1J333~|*20121130
ABC~|*sdaf~|*T1J333~|*20121130
ABC~|*fsdg~|*T1J333~|*20121130
ABC~|*dfsg~|*T1J333~|*20121130

in this i want to count number delimiter occur is 4 in each row if count is less then or more then 4 in any row then it return 1 else 0


my command is

Code:
delim_flag=0
awk -F '~|*' 'BEGIN{delim_flag=0} NF != 4 {delim_flag=1} END{print delim_flag}' "filename.txt"

But this command is not working because delimiter is ~|* same command is working when delimiter is | or ~ or ~| but with * there is problem

Last edited by Don Cragun; 10-26-2015 at 05:56 AM.. Reason: Add CODE and ICODE tags.
# 2  
Old 10-23-2015
Hello Mohanp12,

Welcome to forums, please use code tags for commands/codes/Inputs which you are using in your posts as per forum rules. Following may help you in same.
Code:
awk -F"~|*" 'BEGIN{print "Line Number \t status"} {if(NF>4){print NR OFS OFS 1} else {print NR OFS OFS 0}}' OFS="\t"  Input_file

Output will be as follows.
Code:
Line Number 	 status
1		1
2		1
3		1
4		1

I am putting 1 in output when condition is Number of fields NF>4, while taking delimitor as ~|*. You could change status as per your need too, hope this helps. Enjoy learning.

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
# 3  
Old 10-23-2015
The problem is the | as it has a special meaning in regexes (alternation).
Try
Code:
awk '{print NR, gsub("~\|*", "&")!=3}' file3
1 0
2 0
3 0
4 0

# 4  
Old 10-26-2015
Rabindra

Thanks in advance


The code you provide is given same error . There is issue with *

Code:
[Error]

awk: 0602-521 There is a regular expression error.
        ?*+ not preceded by valid expression.

The input line number is 1. The file is cerf/TGT_OUTBND_20151012_122417DUP.txt.
 The source line number is 1.
DELFLADF=Line Number     status
+ echo Line Number       status
Line Number      status


Last edited by Don Cragun; 10-26-2015 at 02:20 PM.. Reason: Add CODE tags.
# 5  
Old 10-26-2015
Hello MOHANP12,

You haven't mentioned your OS name, on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk , /usr/xpg6/bin/awk , or nawk. Let us know if that helps or please do provide us complete details on your OS and errors for same too. I am using bash(LINUX) system and it is working fine for me.


Thanks,
R. Singh
# 6  
Old 10-26-2015
The asterisk has special meaning in an ERE too.

Assuming you aren't on a Solaris/SunOS system, try:
Code:
awk -F '~\|\*' 'NF != 4 {delim_flag=1; exit} END{printf("%d\n", delim_flag}' "filename.txt"

# 7  
Old 10-26-2015
Hi RavinderSingh13

My OS is AIX unix.

Code:
[Error] 

+ + awk BEGIN{FS="~|*"} {if(NF!=30){print NR OFS OFS 1} else {print NR OFS OFS 0}} OFS=\t TGT_OUTBND_20151012_122417DUP.txt
awk: 0602-521 There is a regular expression error.
        ?*+ not preceded by valid expression.

 The input line number is 1. The file is TGT_OUTBND_20151012_122417DUP.txt.
 The source line number is 1.
DELFLADF=
+ echo


Code:
#!/usr/bin/ksh
set -x
#IFS="~|*" "126 124 42"
DELFLADF=`awk 'BEGIN{FS="~|*"} {if(NF!=30){print NR OFS OFS 1} else {print NR OFS OFS 0}}' OFS="\t" "TGT_OUTBND_20151012_122417DUP.txt"`

#DELFLADF=`nawk -F "~|*" 'BEGIN{delim_flag=0} NF != 30 {delim_flag=1} END{print delim_flag}' "TGT_OUTBND_20151012_122417DUP.txt"`
#DELFLADF=`awk '{print NR, gsub("~\|*", "&")!=30}' "TGT_OUTBND_20151012_122417DUP.txt"`
echo "$DELFLADF"


Already try all condition FS="~|\*" \ for specical chrcter

Last edited by Don Cragun; 10-26-2015 at 06:21 AM.. Reason: Add CODE and ICODE tags; fix other CODE tags.
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