我有xml文件,格式如下,我想用csv转换它,如下所示。不幸的是,我不允许安装xmlstarlet或其他xml解析器(我只有xmllint)。我怎么能这样做呢,比如,呃,……
<xn:VsDataContainer id="site00881">
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008811,ManagedElement=1</es:listOfNe>
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008819,ManagedElement=1</es:listOfNe>
</xn:VsDataContainer>
<xn:VsDataContainer id="site00882">
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008821,ManagedElement=1</es:listOfNe>
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008829,ManagedElement=1</es:listOfNe>
</xn:VsDataContainer>
<xn:VsDataContainer id="site00883">
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=ASDTUR13,MeContext=rbs008831,ManagedElement=1</es:listOfNe>
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=ASDTUR_SIU,MeContext=siu008832,ManagedElement=siu008832</es:listOfNe>
</xn:VsDataContainer>
<xn:VsDataContainer id="site00884">
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008841,ManagedElement=1</es:listOfNe>
<es:listOfNe>SubNetwork=NL1_R,SubNetwork=AHPTUR14,MeContext=rbs008849,ManagedElement=1</es:listOfNe>
</xn:VsDataContainer>输出应采用csv格式
rbs008811,site00881
rbs008819,site00881
rbs008821,site00882
rbs008829,site00882
rbs008831,site00883
siu008832,site00883
rbs008841,site00884
rbs008849,site00884发布于 2015-02-10 16:39:53
我会帮助您处理xmllint,但是您的xml文件并不有效。
总之,这里有一个快速而肮脏的解决方案,您可能应该避免这样做:
grep -Po "(rbs|site)\d+" file.xml | awk '/site/{site=$1} /rbs/{print $1","site}'
rbs008811,site00881
rbs008819,site00881
rbs008821,site00882
rbs008829,site00882
rbs008831,site00883
rbs008841,site00884
rbs008849,site00884发布于 2015-02-10 21:18:22
对于解析XML通常持保留态度:
gawk -v OFS=, '
match($0, /VsDataContainer id="([^"]+)/, m) {container = m[1]}
match($0, /MeContext=([^,]+)/, m) {print m[1], container}
' file如果你没有GNU awk:
awk -v OFS=, '
/VsDataContainer id="/ {
sub(/.*id="/, "")
sub(/".*/, "")
container = $0
}
/MeContext=/ {
sub(/.*MeContext=/, "")
sub(/,.*/, "")
print $0, container
}
' filehttps://stackoverflow.com/questions/28436449
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