Linux

Hello all;

I've been having trouble completing a script (bash) because I can't get past the issue of finding a line in a file. For example, I have a file like this:

ddmmmyyyy Lon Lat
24may2003 -100.0 24.1
25may2003 -100.1 24.0
28may2003 -99.5 23.2
....etc
05jun2003 -97.0 22.7
EOF

I'm making a loop, and I need a way to call a line number without specifying what the line contains (i.e., grep doesn't work). Here's what I had in mind:

n=2
While (something that makes the loop infinite); do
day=`command line${n} word1


if

One way to display a specific line number in a file:

Code:

sed -n 'Xp' input_file

Where "X" is a line number in the "input_file".

Shell_Life, you probably want to quit after the print. There would be no need to read the remainder of the file.

You are right, Reborg -- the quit is an important optmization:

Code:

sed 'X{p;q;}' input_file

Thanks guys, the sed -n command is helpful.

Someone posted another reply, but it got deleted because I double-posted (notice the post is cut-off after "if." I can't remember what happened but I guess I ended up starting over and posted twice accidentally).

Here's jim mcnamara's reply to the deleted post, which got sent to my email:

Code:

awk ' {
     print "set time ", $1
     print "draw mark ", $2, $3
     } ' filename > file_for_a_graphing_program

output:

Code:
---------
set time 24may2003
draw mark -100.0 24.1
set time 25may2003
draw mark -100.1 24.0
set time 28may2003
draw mark -99.5 23.2
set time 05jun2003
draw mark -97.0 22.7
---------
source file = "filename" above:

Code:
---------
24may2003 -100.0 24.1
25may2003 -100.1 24.0
28may2003 -99.5 23.2
05jun2003 -97.0 22.7
---------