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check position of end of line(URGENT PLS)

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# 1  
Old 09-20-2007
check position of end of line(URGENT PLS)
I Have to check in a file that all the lines are ending at same posiotin.

Ex : line 1 is ending at position 88
line 2 should at same position i.e 88

Thanks in advance
# 2  
Old 09-20-2007
Code:
shouldbe=88
awk -v e=$shouldbe '{ if(length($0) != e){print "bad line ", NR} ' filename

# 3  
Old 09-20-2007
Use nawk or /usr/xpg4/bin/awk on Solaris:

Code:
awk 'FNR==1{len=length;next}
length!=len{printf "Check file %s,record %d\n",FILENAME,FNR}' filename[s]


P.S. For portability use length($0).
# 4  
Old 11-08-2007
check position of end of line for some specific lines
Have to check in a file that the lines starting with 620 and 705
are ending at same posiotin.


82012345
62023232323
70523949558
62023255454
9999

In the above lines, i have to check the lines starting with 620 and 705 are having the length 94.

Can anyone help. Immediate response is very much appriciated.
# 5  
Old 11-08-2007
Use the command given above but change the awk to only look at the relevant lines by doing the following:
awk -v e=$shouldbe '{ /^(620|705)/ if(length($0) != e){print "bad line ", NR} ' filename

Should work
# 6  
Old 11-08-2007
check position of end of line for some specific lines
Thankyou very much for your immediate response.

But its giving some syntax error.

this is what i used.

shouldbe=94
srcFile=$1

awk -v e=$shouldbe '{ /^(620|705)/ if(length($0) != e){print "bad line ", NR} }' $srcFile

The error is below
==============
syntax error The source line is 1.
The error context is
{ /^(620|705)/ >>> if <<< (length($0) != e){print "bad line ", NR} }
awk: The statement cannot be correctly parsed.
The source line is 1.


Can you please correct it?
# 7  
Old 11-11-2007
Looks like this was answered for you in another thread:
Quote:
Originally Posted by summer_cherry
Hi,
code:
Code:
awk ' ($0 ~ /^620/ || $0 ~ /^705/ ) && (length($0)==94)' a

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