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how to validate a field when it is blank using awk prog

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Top Forums Shell Programming and Scripting how to validate a field when it is blank using awk prog
# 8  
Old 09-23-2009
Without a sample input file and desired output, it will be difficult to help you.
# 9  
Old 09-23-2009
You could try using the length function

Code:
if (length(f10) == 0){
        print "Field 10 is Correct";}
else{
        print "Field 10 is Wrong";}

# 10  
Old 09-24-2009
its still not working....

I have written the below piece of code for reading a input file which is a fixed length file...

Code:
ls sample.txt | awk '

Code:
  { a[NR]=$1 }
  END{
  FS="n"
  for(i=1;i<=NR;i++)
     {
            while( getline < a[i] ) 
                        {
                                       f1=$0;
                                       print("Line::",f1);  
                        f2=substr(f1,1,1)     
                        print("Field1::",f2);
                        f3=substr(f1,2,1)
                        print("Field2::",f3);
                        f4=substr(f1,3,6)
                        print("Field4::",f4);
                        f5=substr(f1,9,2)     
                        print("Field5::",f5);
                        f6=substr(f1,11,58)
                        print("Field6::",f6);
                        f7=substr(f1,69,17)     
                        print("Field7::",f7);
                        f8=substr(f1,86,11)
                        print("Field8::",f8);
                        f9=substr(f1,97,1)     
                        print("Field9::",f9);
                        f10=substr(f1,98,7)
                        print("Field10::",f10);
                        if(length(f10) == 0){
                        print "Field 10 is Correct";}
                        else{
                         print "Field 10 is Wrong";}
                        f11=substr(f1,105,21)     
                        print("Field11::",f11);
                        f12=substr(f1,126,15)
                        print("Field12::",f12);  
                        f13=substr(f1,141,15)
                        print("Field12::",f13);         
                        f14=substr(f1,156,2)
                        print("Field13::",f14); 
                        if(f14 ~ /^["][ ]+["]$/){
                         print "Field 13 is Correct"}
                        else{
                         print "Field 13 is Wrong"}
                        f15=substr(f1,158,2)
                        print("Field14::",f15); 
                                    } 
              }
     }'



Quote:
sample.txt--->name of the input file
Quote:
content of the file will be specified in the below format
0000000200aaaaaa aaaaaaaa aaaaaaaaa 00/00/00 00:00:00000000 1 1111111 2222.22 2222.22 00


the output for the above input is
Quote:
Line:: 0000000000aaaaaa aaaaaaaa aaaaaaaaa 00/00/00 00:00:00000000 1 1111111 2222.22 2222.22 00
Quote:

Field1:: 0
Field2:: 0
Field4:: 00000
Field5:: 00
Field6:: aaaaaa aaaaaaaa aaaaaaaaa
Field7:: 00/00/00 00:00:00
Field8:: 000000
Field9:: 1
Field10::
Field 10 is Wrong
Field11:: 111111
Field12:: 2222.22
Field12:: 2222.22
Field13::
Field 13 is Wrong
Field14:: 00


But instead of displaying it as Correct it prints wrong.

Please help me on knowing what went wrong in the above piece of code
# 11  
Old 09-24-2009
If I understand, you are expecting a space in field 10 and 13 right?

If your check fails it is probably because there are more spaces than you expect. Try printing them with delimiters to check the output:

Code:
printf "Field10::>%s<", f10;



---------- Post updated at 11:09 AM ---------- Previous update was at 11:00 AM ----------

I can already see your problem:
Code:
f10=substr(f1,98,7)
print("Field10::",f10);
if(f10 ~ /^[[:space:]]$/){
	print "Field 10 is Correct"}
else{
	print "Field 10 is Wrong"
}

For f10 you extract 7 characters from f1. The next test will never match on single space.
# 12  
Old 09-24-2009
we cannot add any delimiter to that file for printing the field values and it is a fixed length file

here what we expect is that we have 7 characters which is expected to be blank even if one of them is not then we need to have get an error message.

Please help us in this.

Thanks in advance
meva
# 13  
Old 09-25-2009
Quote:
Originally Posted by meva
we cannot add any delimiter to that file for printing the field values and it is a fixed length file

here what we expect is that we have 7 characters which is expected to be blank even if one of them is not then we need to have get an error message
That's what I thought. You are trying to match a string with 7 space characters against the pattern ^[[::space::]]$ which means exactly *one* space character.

Try the pattern ^[[::space::]]+$.

As for the delimiter, my suggestion is only a testing trick to visualize the length of a string of spaces.

What version of awk are you using? For fixed length data, GNU awk as the FIELDWIDTHS built-in variable that facilitate the parsing. For other versions of awk there is a workaround. Let me know.

---------- Post updated at 09:00 AM ---------- Previous update was at 08:51 AM ----------

Also, there is a inconsistency in your data for field 10 and 13. Your substr() function extracts respectively 7 and 15 characters that are supposed to be spaces but the sample you provides only shows one single space at that place. Please check.
# 14  
Old 09-29-2009
Hi Ripat,

Thanks for your help.

I got the solution when i tried to check the space of the field with the below pattern
Quote:
/^\ *$/
Meva
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