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If with grep in awk

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# 1  
Old 09-17-2014
If with grep in awk
Hi,

I have i/p file below

Code:
xxx.com
server version
dfsdfdfsd
asdafsf
---end---
diskspace
zdcdfzxcx
ZXccxvc

yyy.com
server version
w4ewr0as0dias90
324rwefdsok0
---end---
diskspace
.......


ddd.com
server version

ttt.com
diskspace

have four set of data, the first line is server name .com highlighted in blue.

if the server name found then it has to check server version (highlighted in red) and disk space (highlighted in purple) if both version and disk space present redirect to o/p1.txt file

if only disk space present redirect to another file o/p2.txt

if only server version redirect to other file o/p3.txt


below is my code just sample

Code:
awk ' 
a='server varsion' & b='disksapce'
{if (/.com/== 0) print $a $b > o/p.txt
}'  inputfile.txt


so if both disksapce andserver version present all the content (junk under server version "dfsdfdfsd asdafsf") should move to o/p file. shall i use
Code:
getline;print

to pint next line.

so please help me whether my code will work
Last edited by stew; 09-17-2014 at 07:16 AM..
# 2  
Old 09-17-2014
There are some syntax errors.
You cannot use the / character as part of filename - it means a directory.
Code:
a='server varsion' && b='disksapce'
awk -v a="$a" -v b="$b" '  /.com$/ {next}  {print a, b >"output.txt"} '  inputfile.txt

I've corrected syntax. The logic makes no sense to me. Please give expected output samples.
# 3  
Old 09-17-2014
I have a file like below (five set of data)

Code:
Input file.txt

aaaa
server-version
dsfsrjjgorgk
sfoweiurtit
disk space
ldkosdife
dsfjioeufs

bbbb
server-version
fker9i5toek
dsfmeu90e
disk space
fsgter6y
sfe5t5gf

cccc
server-version
sdt4wtrei
sfd5tet

dddd
disk space
ssretr
dferte56y

eeee
server-version
.............

my requirement is need to pass all server name one by one (aaaa,bbbb,cccc,ddddd......)

once we pass server name need to check both server-version and disk space are available if yes then redirect to o/p file like below

Code:
output1.txt


aaaa
server-version
dsfsrjjgorgk
sfoweiurtit
disk space
ldkosdife
dsfjioeufs

bbbb
server-version
fker9i5toek
dsfmeu90e
disk space
fsgter6y
sfe5t5gf

If only server-version is available rediect another file

Code:
output2.txt

cccc
server-version
sdt4wtrei
sfd5tet

eeee
server-version
.............

If only disk space is available redirect to third o/p file.

I am not sure how we can pass server name one by one (aaaa,bbbb,cccc..........) since server names are different.

so once I can pass it I will assign server-version and disk space in variable.
Last edited by stew; 09-17-2014 at 02:46 PM..
# 4  
Old 09-17-2014
Store all hostnames in a file ("servers" for example)...
Code:
cat servers
aaaa
cccc
dddd

Run the awk script on "InputFile.txt" to get the desired output separated by criteria into different files...
Code:
awk 'BEGIN {while (getline l < "servers" > 0) s[l]; RS=""; FS="\n"}
{
    if ($1 in s) {
        if ($0 ~ "server-version" && $0 ~ "disk space")        print $0 > "output1.txt"
        else if ($0 !~ "server-version" && $0 ~ "disk space")  print $0 > "output2.txt"
        else if ($0 ~ "server-version" && $0 !~ "disk space")  print $0 > "output3.txt"
    }
}' InputFile.txt

This User Gave Thanks to shamrock For This Post:
stew
# 5  
Old 09-17-2014
but is there any way without placing all server name in a separate file to pass value
# 6  
Old 09-17-2014
Looks like you have a blank line between each record, so awk can split records on a blank line:

Code:
awk '/server-version/ && /disk *space/ {print $0 "\n"}' RS= inputfile.txt

Also note that I put " *" in between disk and space, as I'm unsure if you datafile has a space there or not.
# 7  
Old 09-18-2014
Thanks ShamRock

My doubt is if I pass only server name in output file 0
Code:
BEGIN {while (getline l < "servers name input file" > 0)

Its has only server (aaaa,bbbb,cccc...) then I can compare server name with server-version
Code:
aaaa="server-version

i.e,  if ($0 ~ "server-version" && $0 ~ "disk space")        print $0 > "output1.txt"
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