用 xmlstarlet 迭代 xml 并输出父节点和子节点值

Question

我在多个 XML 文件中有这种格式：

<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.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>

我想要这样的输出：

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"

我有一个 bash 脚本，它使用 xmlstarlet 遍历目录中的 xml 文件，但它会转储最后一个 desc 之后的所有 "related values" ID。它需要将每个 desc id 与每组 "related" 值相关联。并且需要包含每个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

Answer 1

#!/bin/bash

for x in *.xml; do
  count=$(xml sel -t -v 'count(//bad/objdesc/desc/@id)' "$x")
  for ((i=1; i<=count; i++)); do
    id=$(xml sel -t -v "//bad/objdesc/desc[$i]/@id" "$x")
    arr=( $(xml sel -t -v "//bad/objdesc/desc[$i]/related/@objectid" "$x") )
    cat<<EOF
$id related="$(perl -e 'print join ",", @ARGV' "${arr[@]}")"
EOF
  done
done

=)

这似乎是 XSLT 的工作。但是，好的，shell 也可以处理这个...

你能为 dbi 完成剩下的工作吗？最好尝试理解这里涉及的内容，而不仅仅是 cut/paste.

Answer 2

同意 sputnick 的观点，XSLT 是正确的工具。尽管如此，使用 XML 标记解析器的 perl 答案。优点是它只需要处理一次文件而不是重复调用xmlstarlet:

#!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：

$ 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"

Answer 3

$ xml sel -t -m bad/objdesc/desc -v "concat(@id,' dbi=',@dbi,' ')" -m related -v @objectid -i "number(count(./preceding-sibling::related))+1<number(count(./../related))" -o ", " --else -n -b file.xml

butwba10.1.wc.01 dbi=BUTWBA10.1.1.WC 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 but621r.1.penc.01, bb435.1.comdes.03

用 xmlstarlet 迭代 xml 并输出父节点和子节点值

iterate over xml with xmlstarlet and output parent and child node values

xml

bash

xmlstarlet