自动提取Word docx存档和漂亮的xml文件的打印转换

Question

如果重命名例如文件文档.docx为document.docx.unzipped.zip则可以提取那个存档，例如。到文件夹">文档.docx.解压缩"。不幸的是，提取的xml文件不是很好读，因为所有xml信息都在一行中。

我想自动化提取docx存档并将所有xml文件从存档或提取文件夹(文档.docx.解压缩(转换为可读/漂亮的打印版本(如记事本++-->扩展-->XML工具-->漂亮的打印(XML仅带有换行符((

有什么快速方法的想法吗？

编辑1：从 https://stackoverflow.com/users/1761490/pawel-jasinski 修改的想法

#!/bin/sh

# this scripts unpacks and reformat docx files
#
# you need xslt processor (Transform) in your path
# /c/Program Files/Saxonica/SaxonHE9.4N/bin/Transform
#
# make sure to copy remove-rsid.xslt and copy.xslt
if [ "$1" = "-r" ]; then
    remove_rsid=1
    shift
fi
if [ "$1" = "" ]; then
    echo expected name of the word document to be exploded
    exit 1
fi
suffix=${1##*.}
name="$1"
if [ "$suffix" = "xml" ]; then
    suffix=docx
    name=${1/%.xml/.docx}
fi
if [ "$suffix" = "$1" ]; then
    suffix=docx
    name=$1.docx
fi

corename=$(basename "$name" .$suffix)
if [ -z "$corename" ]; then
    echo can not work with empty name
    exit 1
fi
DIR="$( cd "$( dirname "$0" )" && pwd )"
DOSDIR=$(cygpath -m $DIR)
FLAT=$PWD/$corename.tmp/flat.$$
FLATOUT=$PWD/$corename.tmp/flat.$$.out

if [ "$remove_rsid" == "1" ]; then
    transform=$DOSDIR/remove-rsid.xslt
else
    transform=$DOSDIR/copy.xslt
fi
# $1 - file name
# 
# formats file as xml
_reformat_xml() {
    echo reformat $1
    #read pause
    xmllint --format $1 -o $1.new
    mv $1.new $1
}
flaten() {
    # xml
    xmls=""
    pwd
    pwd
    #read pause
    for f in $(find . -name '*.xml'); do  
        ff=$(echo ${f#./} | tr '/' '@')
        echo mv $f $FLAT/$ff
        mv $f $FLAT/$ff
        _reformat_xml $FLAT/$ff
        xmls="$xmls $ff"
    done
    # for rels, rename into .xml
    rels=""
    for f in $(find . -name '*.rels'); do  
        ff=$(echo ${f#./} | tr '/' '@')
        rels="$rels $ff"
        mv $f $FLAT/$ff.xml
        _reformat_xml $FLAT/$ff.xml
        #read pause
    done
}
expand_dirs() {
    target_dir=$(pwd)
    cd $FLATOUT
    echo PDW: $PWD
    #read pause
    for f in $rels ; do
        ff=$(echo $f | tr '@' '/')
        mv $f.xml "$target_dir/$ff"
    done
    for f in $xmls ; do
        echo PDW: $PWD
        #read pause
        ff=$(echo $f | tr '@' '/')
        mv $f "$target_dir/$ff"
    done
    cd "$target_dir"
}
echo corename: $corename
read pause
if [ -e "$corename" ]; then
    if [ -e "$corename.bak" ];then
        # echo removing $corename.bak
        rm -rf "$corename.bak"
    fi
    # echo backing up $corename
    mv "$corename" "$corename.bak"
fi 

mkdir "$corename"
cd "$corename"
unzip -q "../$name"
if [ -a $FLAT ]; then
    rm -rf $FLAT
fi
mkdir -p $FLAT
flaten
if [ -a $FLATOUT ]; then
    rm -rf $FLATOUT
fi
mkdir -p $FLATOUT
#exit
#dosflat=$(cygpath -m $FLAT)
#Transform -xsl:$transform -s:$dosflat -o:$dosflat.out
cp -R $FLAT/* $FLATOUT
expand_dirs
read pause #
rm -rf $FLAT $FLATOUT

Answer 1

如果你曾经使用过cygwin，它包括xmllint而又具有--format选项。这是我最初的做法。但是xmllint没有按照我喜欢的方式格式化属性，所以我开发了自己的脚本。由于word文档包含大量rsid干扰，因此脚本可以选择将其删除。

我使用以下工作流程：

• 获取一个 Word 文档，比如说foo.docx
• explode-docx -r foo.docx
• 编辑foo.docx - 进行小改动
• explode-docx -r foo.docx
• kdiff3 foo foo.bak