Question

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.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>

我希望输出看起来像这样：

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迭代目录中的xml文件，但它会在最后一个desc id之后转储所有“相关值”。它需要将每个desc id与每组“相关”值相关联。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

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，好吧，shell也可以处理这个......

你能为dbi完成剩下的工作吗？尝试理解这里涉及的内容比剪切/粘贴更好。

Answer 2

同意sputnick XSLT是正确的工具。然而，使用XML令牌解析器的perl答案。它的优点是它只需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]);

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"

使用xmlstarlet迭代xml并输出父节点和子节点值

2 个答案: