使用xmlstarlet在xml上迭代并输出父节点和子节点值
我在多个XML文件中使用此格式:使用xmlstarlet在xml上迭代并输出父节点和子节点值,xml,bash,xmlstarlet,Xml,Bash,Xmlstarlet,我在多个XML文件中使用此格式: <bad> <objdesc> <desc id="butwba10.1.wc.01" dbi="BUTWBA10.1.1.WC"> <physdesc>adfa;sdfkjad</physdesc> <related objectid="bb435.1.comdes.02"/> <related objectid="but614r.
<bad>
<objdesc>
<desc id="butwba10.1.wc.01" dbi="BUTWBA10.1.1.WC">
<physdesc>adfa;sdfkjad</physdesc>
<related objectid="bb435.1.comdes.02"/>
<related objectid="but614r.1.penc.01"/>
<related objectid="but611.1.wc.01"/>
<related objectid="but612.1.wd.01"/>
<related objectid="bb515.1.comb.12"/>
</desc>
<desc id="butwba10.1.wc.02" dbi="BUTWBA10.1.2.WC">
<physdesc>alkdjfa;sfjsdf</physdesc>
<related objectid="but621r.1.penc.01"/>
<related objectid="bb435.1.comdes.03"/>
</desc>
</objdesc>
</bad>
我有一个bash脚本,它使用xmlstarlet迭代目录中的xml文件,但它会转储最后一个desc id之后的所有“相关值”。它需要将每个desc id与每一组“相关”值关联起来。它需要包含每个id的dbi值
#!/bin/bash
for x in *.xml
do
id=$(xml sel -t -v '//bad/objdesc/desc/@id' "$x")
arr=( $(xml sel -t -v '//bad/objdesc/desc/related/@objectid' "$x") )
cat<<EOF >> new_file
$id related="$(perl -e 'print join ",", @ARGV' "${arr[@]}")"
EOF
done
#/bin/bash
对于*.xml中的x
做
id=$(xmlsel-t-v'//bad/objdesc/desc/@id''$x)
arr=($(xml sel-t-v'//bad/objdesc/desc/related/@objectid''x'))
cat新文件
$id related=“$(perl-e'打印连接',”,@ARGV'${arr[@]}”)
EOF
完成
#/bin/bash
对于*.xml中的x;做
count=$(xmlsel-t-v'count(//bad/objdesc/desc/@id)“$x”)
for((i=1;i同意sputnick的观点,XSLT是正确的工具。不过,使用XML令牌解析器的perl答案具有只需处理一次文件而不必重复调用xmlstarlet的优点:
#!perl
use strict;
use warnings;
use XML::Parser;
my (@related, @desc); # boo, global variables
sub start {
my ($x, $elem, %attrs) = @_;
if ($elem eq "desc") {
@desc = @attrs{'id', 'dbi'};
@related = ();
}
elsif ($elem eq "related") {
push @related, $attrs{objectid};
}
}
sub end {
my ($x, $elem) = @_;
if ($elem eq "desc") {
printf qq{%s dbi="%s" related="%s"\n}, @desc, join(', ', @related);
}
}
my $parser = XML::Parser->new( Handlers => {Start => \&start, End => \&end} );
$parser->parsefile($ARGV[0]);
在行动中:
$ perl parse.pl file
butwba10.1.wc.01 dbi="BUTWBA10.1.1.WC" related="bb435.1.comdes.02, but614r.1.penc.01, but611.1.wc.01, but612.1.wd.01, bb515.1.comb.12"
butwba10.1.wc.02 dbi="BUTWBA10.1.2.WC" related="but621r.1.penc.01, bb435.1.comdes.03"
很好,但是我们有3个调用xmlstarlet
。这不是最优的,但它是有效的:)调用perl连接的另一个替代方法是join=$(IFS=,;echo“${arr[*]}”)
——仍然需要生成子shell思想,或者在当前shell中,printf-v join”%s,“${arr[@}”;join=${join%,}
三次使用xmlstarlet的原因是,在SO的另一篇帖子中,要求不够明确。事情一点一点地发生了=)>因为需要时间才能给出正确的解决方案+1
#!perl
use strict;
use warnings;
use XML::Parser;
my (@related, @desc); # boo, global variables
sub start {
my ($x, $elem, %attrs) = @_;
if ($elem eq "desc") {
@desc = @attrs{'id', 'dbi'};
@related = ();
}
elsif ($elem eq "related") {
push @related, $attrs{objectid};
}
}
sub end {
my ($x, $elem) = @_;
if ($elem eq "desc") {
printf qq{%s dbi="%s" related="%s"\n}, @desc, join(', ', @related);
}
}
my $parser = XML::Parser->new( Handlers => {Start => \&start, End => \&end} );
$parser->parsefile($ARGV[0]);
$ perl parse.pl file
butwba10.1.wc.01 dbi="BUTWBA10.1.1.WC" related="bb435.1.comdes.02, but614r.1.penc.01, but611.1.wc.01, but612.1.wd.01, bb515.1.comb.12"
butwba10.1.wc.02 dbi="BUTWBA10.1.2.WC" related="but621r.1.penc.01, bb435.1.comdes.03"