Shell Programming and Scripting

i have a log file while looks like this

++
user_a blabla
blabla nas_b blabla user_d
this is a user_a
junk line
another junk line
user_c nas_m blabla
++

basically most of the lines contain a "user" keywords, and the rest of the lines do not have "user" at all.

So I have the following script, which just extracts the user part(if that line has a user keyword) from each line

awk '{ for (i=0; i<=NF; i++)
if($i~/user/)
{a=$i}
else
{continue}
{print a}
}' /tmp/test

and when I run it, I got the following result

user_a
user_d
user_a
user_a #<===actually there is no user in that line at all.
user_c

why this happens and how can I fix my script? thanks

I think it is because you are always printing a value. So, you are not over-writing 'a' ni the instance of the 4th line, so when you print the variable, you are getting the last stored value of 'a'.
Perhaps as the first line ater your else, do a

Code:

a=""

to blank out the last 'a' value?

I reliazed that problem, but it seems I can not fix the problem, here is the modified script

awk '{ for (i=0; i<=NF; i++)
if($i~/user/)
{a=$i}
else
{a="no"}
{ continue}
{print a}
}' /tmp/test

and the output change to

no
user_d
user_a
no
no

which is not correct at all.

With awk:

Code:

awk '{ 
  for (i=1; i<=NF; i++)
    if ($i ~ /user/)
      print $i
      }' infile

With Perl:

Code:

perl -lne'print $1 while /(user[^\s]*)/g' infile

With GNU grep:

Code:

grep -o 'user[^ \t]*' infile

Thanks, I know in this example I gave, egrep is a better choice, but actually my real work is "finding out both "user" and "nas" part, and then print out the result.

below is the result I want

+
user_a nas_b
user_d nas_b
user_a
user_c nas_m
+

But the problem is, no matter how i modified my script, I just won't get the desired result...

[QUOTE=radoulov;302286540]With awk:

Code:

awk '{ 
  for (i=1; i<=NF; i++)
    if ($i ~ /user/)
      print $i
      }' infile

With Perl:

I don't understand the logic ...
Why user_d nas_b?

sorry, that was a typo