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Print records which do not have expected number of fields in a comma delimited file

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# 1  
Old 01-31-2012
Print records which do not have expected number of fields in a comma delimited file
Hi,

I have a comma (,) delimited file, in which few fields are enclosed with in double quotes " ". I have to print the records in the file which donot have expected number of field with the line number.

Code:
 
File1
====
name,desgnation,doj,project #header#
asath,se,15-06-2010,asc,"india,mumbai"
rif,sse,12-05-2010,asc
si,tl,01-12-2009,asc
mthr,"ase,trans",15-09-2010
sdu,ase,15-09-2010,bench
 
Here lets say me expected number of field is 4. And in the above input, record 2 has 5 fields and record 5 as 2 fields. I have to print the two records with line number.
 
Expected output
 
File1_fldmismatchrcrds
==================
2 asath,se,15-06-2010,asc,"india,mumbai"
5 mthr,"ase,trans",15-09-2010

I tried the below code

Code:
 awk -F "," '{gsub(/"[^"]*"/,x);if (NF != '4'){print NR,$0}}' File1

This gives me the output as
Code:
2 asath,se,15-06-2010,asc,
5 mthr,,15-09-2010



Can someone please help on this issue?
Smilie
# 2  
Old 01-31-2012
awk -F "," '/"/{print NR,$0}' file
This will help you.

Output-:
2 asath,se,15-06-2010,asc,"india,mumbai"
5 mthr,"ase,trans",15-09-2010
# 3  
Old 01-31-2012
Hi Could you please explain how this works?
This returns me the rows in which there are double quoted (" ") but what I require is I have to print any record (with / without double quotes enclosed data) in it if the number of field is less than expected number of field.
The above code prints the records with double quote enclosed data alone. Irrepective of the number of fields.
# 4  
Old 01-31-2012
Try preserving the record first:
Code:
awk -F, '{p=$0;gsub(/"[^"]*"/,x)} NF!=4{print NR,p}' file

# 5  
Old 01-31-2012
Code:
/pattern/ {statements}

is a pattern match and the function or series of statments to invoke if the pattern matches
ie., a boolean result (true or false)

Code:
# expected number is 4
# this prints the record number if there are fewer than 5 fields
# and if there is a quote somewhere on the line
awk -F ','   ' /"/ && NF!=4 {print NR,$0}' file

I do not get exactly what the criteria for printing are supposed to be. So this is just giving you what it seems you want....
# 6  
Old 01-31-2012
Thanks. It worked!! Smilie
Code:
awk -F, '{p=$0;gsub(/"[^"]*"/,x)} NF!=4{print NR,p}' file

Last edited by machomaddy; 01-31-2012 at 09:25 AM..
# 7  
Old 01-31-2012
Hi machomaddy
Are you not getting the output what you want. If not please show me the output you want
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