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match pattern 1 and print or match pattern 2 and print

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# 1  
Old 03-03-2011
match pattern 1 and print or match pattern 2 and print
hi all

basically i have file called rules which contain lines like below

Code:
/usr/bwmgr/utils/bwmgr em1 -x 735 -name user92 -addr 10.10.201.92 -addrmsk 255.255.255.252 -bwout 1024000 -bwin 2048000 -statsdevice user92 -stats
/usr/bwmgr/utils/bwmgr em1 -x 45032 -name user246 -addr 10.10.224.246 -bwboth 128000 -bwlink grp1 -statsdevice user246 -stats


i have a script which basically looks for the rule number and prints out the bandwidth (in and out)

Code:
sbpc# cat script.sh
#!/bin/sh
RuleNum=$1

/bin/cat /usr/home/bw/rules | /usr/bin/sed 's/^.*-x //' | /usr/bin/awk -v var=$RuleNum '$1==var {for(i=1;i<=NF;i++) {if($i=="-bwout") print $(i+3),$(i+1)}}'
sbpc#

so if i run the script and give input as 735, it will output

2048000 1024000

Code:
sbpc# sh script.sh 735
2048000 1024000
sbpc#

but if I do "sh script.sh 45032" then nothing will be printed out because the script cannot match the pattern "-bwout".
the file lines either contain the word "-bwout" or the word "-bwboth".
how can i change the script so that it will give either match "-bwout" and give output like its doing now or it will match "-bwboth" and print the next word.

so that if i do "sh script.sh 45032" then result should be 128000 .

with the same script i want to get output regardless of pattern "-bwout" or "-bwboth"

tia
# 2  
Old 03-03-2011
Hi sb245, Try this,
Code:
/bin/cat /usr/home/bw/rules | /usr/bin/sed 's/^.*-x //' | /usr/bin/awk -v var=$RuleNum '$1==var {for(i=1;i<=NF;i++) {if($i=="-bwout" ){ print $(i+3),$(i+1)}if($i=="-bwboth"){print $(i+1)}}}'

# 3  
Old 03-03-2011
Maybe you can start from this code:

Code:
grep -Eo "\-bwout [^ ]*|\-bwin [^ ]*" infile

-bwout 1024000
-bwin 2048000

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