check position of end of line(URGENT PLS)

Shell Programming and Scripting
1

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 
shouldbe=88
awk -v e=$shouldbe '{ if(length($0) != e){print "bad line ", NR} ' filename
3

Use nawk or /usr/xpg4/bin/awk on Solaris:

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

P.S. For portability use length($0).

4

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

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

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

Looks like this was answered for you in another thread: