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how to fix this awk script?

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Top Forums Shell Programming and Scripting how to fix this awk script?
# 8  
Old 02-11-2009
an actual (with blah-s) file sample would be helpful. A desired output (based on a sample input) would not hurt either.
Please use vB Codes when posting samples/quotes.
# 9  
Old 02-11-2009
sure, let me be more clear, here is the file test

$more /tmp/test
user_a blabla nas_b
blabla nas_b blabla user_d
this is a user_a
another junk line
user_c nas_m blabla

please note, user and nas may or may not appear on each line, and the position of user and nas are not fixed, they can appear in any column.

I need a script to scan each line, if any line contains user or nas, or both, they will be extracted from the line and be printed out, and i want print user before nas if a line has both words. in my case, the result will be look like:

user_a nas_b
user_d nas_b
user_a
user_c nas_m

I tried to monky with awk for a while, but just can not figure out a good way to make it.

Any help will be greatly appreciated.
# 10  
Old 02-11-2009
Code:
awk  '/user|nas/{ for (i=1; i<=NF; i++) {
                   if ($i ~ /user/) printf $i FS 
                   if ($i ~ /nas/) a[i]=$i } 
                  for (j in a) printf a[j] FS
                  print "" ; split("",a) }' filename

# 11  
Old 02-11-2009
this script works for my example ! I must admit I don't understand this awk array thing, I figure there are tons of awk stuff to learn.

but the thing is, it appears that this script does not always work , for example, when I run the script against the following line, i got error

$more /tmp/test2
08:05:55.255412 rad-access-req 58 [id 57] Attr[ Service_type{Authenticate Only} user{Defcon} Pass nas_ipaddr{1.2.2.3} ] (DF)

and the error message is:

"awk: a is not an array
record number 1"


Quote:
Originally Posted by rubin
Code:
awk  '/user|nas/{ for (i=1; i<=NF; i++) {
                   if ($i ~ /user/) printf $i FS 
                   if ($i ~ /nas/) a[i]=$i } 
                  for (j in a) printf a[j] FS
                  print "" ; split("",a) }' filename

# 12  
Old 02-11-2009
Given your requirements, the code works fine either with gawk or old awk. Try to modify it to get your desired output.

Code:
$ cat file
user_a blabla nas_b
blabla nas_b blabla user_d
this is a user_a
another junk line
08:05:55.255412 rad-access-req 58 [id 57] Attr[ Service_type{Authenticate Only} user{Defcon} Pass nas_ipaddr{1.2.2.3} ] (DF)
another junk line
user_c nas_m blabla

$ awk  '/user|nas/{ for (i=1; i<=NF; i++) {
                   if ($i ~ /user/) printf $i FS 
                   if ($i ~ /nas/) a[i]=$i } 
                  for (j in a) printf a[j] FS
                  print "" ; split("",a) }' file

user_a nas_b
user_d nas_b
user_a
user{Defcon} nas_ipaddr{1.2.2.3}
user_c nas_m

# 13  
Old 02-11-2009
after I install gawk, problem fixed, thanks a lot!
Last edited by fedora; 02-12-2009 at 04:12 PM..
# 14  
Old 02-12-2009
sorry, one more question, I have been trying to read some awk documents, but apparently the time is just not enough for me to pick up the awk "array" very quickly. Right now I just saved the script.

If I insert one more keyword group into the test file, how can i tune this awk script to pick up all 3 keywords and print user first, nas second and group last?

exmple,
$more /tmp/test
user_a blabla nas_b group_a
blabla nas_b group_b blabla user_d
this is a user_a, group_c
another junk line
group_d user_c nas_m blabla

and the desire output

user_a nas_b group_a
user_d nas_b group_b
user_a group_c
user_c nas_m group_d

thank in advance SmilieSmilieSmilie!
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