[英]Extracting XML elements which contains a certain string with sed
我有一个像下面的文件
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show databases"/>
<AUDIT_RECORD TIMESTAMP="2013-07-29T17:27:53" NAME="Quit" CONNECTION_ID="12" STATUS="0"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show grants for root@localhost"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="create table stamp like paper"/>
在这里,每个记录都以<AUDIT_RECORD
开头,以"/>
结尾,并且该记录可能分布在多行中。
我的要求是显示如下结果
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show databases"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show grants for root@localhost"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="create table stamp like paper"/>
为此,我用了
sed -n "/Query/,/\/>/p" file.txt
但是它将显示整个文件,包括带有字符串“ Quit”的记录。
有人可以帮我吗? 另外请让我知道是否可以完全匹配名为“ Query”的字符串(例如grep -w "Query"
)。
使用GNU awk,您可以将RS设置为多个字符:
$ cat file
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query"
CONNECTION_ID="10" STATUS="0" SQLTEXT="show databases"/>
<AUDIT_RECORD TIMESTAMP="2013-07-29T17:27:53"
NAME="Quit" CONNECTION_ID="12" STATUS="0"/>
<AUDIT_RECORD
TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10"
STATUS="0" SQLTEXT="show grants for root@localhost"/>
<AUDIT_RECORD
TIMESTAMP="2013-07-30T17:52:29"
NAME="Query"
CONNECTION_ID="10"
STATUS="0"
SQLTEXT="create table stamp like paper"/>
$
$ gawk -v RS='\\/>\n' -v ORS= '/Query/{print $0 RT}' file
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query"
CONNECTION_ID="10" STATUS="0" SQLTEXT="show databases"/>
<AUDIT_RECORD
TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10"
STATUS="0" SQLTEXT="show grants for root@localhost"/>
<AUDIT_RECORD
TIMESTAMP="2013-07-30T17:52:29"
NAME="Query"
CONNECTION_ID="10"
STATUS="0"
SQLTEXT="create table stamp like paper"/>
$
$ gawk -v RS='\\/>\n' -v ORS= '/Query/{$1=$1; print $0 RT}' file
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show databases"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="show grants for root@localhost"/>
<AUDIT_RECORD TIMESTAMP="2013-07-30T17:52:29" NAME="Query" CONNECTION_ID="10" STATUS="0" SQLTEXT="create table stamp like paper"/>
我同意@choroba的观点,即XML解析器是正确的工具。 但是,如果没有可用的,则可以尝试以下awk脚本:
awk '/Query/{print RS" "$0}' RS='<AUDIT_RECORD' file
输入的内容可能是XML。 使用适当的解析器来处理它,尤其是当记录跨越多行时。 例如, xsh :
open file.xml ;
remove //AUDIT_RECORD[not(@NAME="Query")] ;
save :b ;
我建议的sed解决方案:
sed 's/<[^>]*\"Quit\"[^>]*>//' file.txt
对于跨越多行的记录,请尝试:
sed '{:q;N;s/\n/ /g;t q}' file.txt | sed 's/<[^>]*\"Quit\"[^>]*>//'
添加换行符RS:
... | sed 's|/>|/>\n|g'
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