[英]How do I extract words between matching characters and combine them into new lines in awk/sed?
知道如何從這些響應中刪除所有格式並僅保留用逗號分隔的“值”部分嗎? 我已經實現了類似的功能,但是需要多次運行並使用單獨的腳本。
response:
{
type => 'query'
timestamp => '1444304880'
serial => '0000012970'
address => '192.168.1.1'
profile => 'common'
query-id => '001'
flags => '(NET, CORP)'
version => '1.0.0.3'
}
response:
{
type => 'query'
timestamp => '1444305643'
serial => '0000012971'
address => '192.168.1.2'
profile => 'common'
query-id => '002'
flags => '(CORP)'
version => '1.0.0.3'
}
理想輸出:
query, 1444304880, 0000012970, 192.168.1.1, common, 001, (NET, CORP), 1.0.0.3
query, 1444305643, 0000012971, 192.168.1.2, common, 002, (CORP), 1.0.0.3
我注意到我可以采用兩種方法,第一種方法是簡單地打印出$ 3和$ 4列:awk'{print $ 3,$ 4}'dump.txt這給了我:
'query'
'1444304880'
'0000012970'
'192.168.1.1'
但是它也包括由“ {}”創建的空格,我可以消除這些空格。 我的另一種選擇是采用這種方式,取出響應{}部分。
sed "s/\'//g" dump.txt | awk '/\{/{flag=1;next}/\}/{flag=0}flag'
但是然后我必須使用以下命令調出每一行:
sed -e '/type/{N;s/\n//;}'
感謝以更好的方式執行此操作。
使用GNU grep並粘貼:
grep -Po "=> '\K.*(?=')" file | paste -d , - - - - - - - -
輸出:
query,1444304880,0000012970,192.168.1.1,common,001,(NET, CORP),1.0.0.3 query,1444305643,0000012971,192.168.1.2,common,002,(CORP),1.0.0.3
假設您的所有輸入都按照示例中的結構進行構建,那么這應該可以工作:
cut -d '>' -f 2 foo.txt | grep "^ " | paste -d, - - - - - - - - | tr -d "'" | sed 's/^ //'
awk:使用“ =>”作為字段分隔符
awk -F "=>" '
# a line with 2 fields, remove single quotes and print with a comma
NF == 2 {gsub(/\x27/, "", $2); printf "%s,", $2}
# end of record, overwrite the trailing command and add a newline
$0 == "}" {printf "\b \n"}
' file
使用gnu-awk
和FPAT
另一種解決方案
awk -vFPAT="['][^\n]+[']" -vRS="{" -vOFS="," '
NR>1{$1=$1; gsub(/\x27/,""); print}' file
你得到,
query,1444304880,0000012970,192.168.1.1,common,001,(NET, CORP),1.0.0.3 query,1444305643,0000012971,192.168.1.2,common,002,(CORP),1.0.0.3
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