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需要将数据从xml解析为shell脚本

tag-icon • tag-icon shell-script xml xmllint
需要将数据从xml解析为shell脚本

需要将数据从xml逐一解析为shell脚本

Requriment :- we need to take <FileSetfolder> as one record. XML may contain many <FileSetFolder> from Each <FileSetFolder> record. 
we need to extract <path> and <Filetypes> and archive all the files which are with extension type as <Filetypes> 
need help in approach for this requirement

ex:- for first <FileSetfolder>
    Archive_path = D:\apache-2.4.10\htdocs
    Filetypes =(rep,zip,mnt)

    need to archive files in $Archive_path which have $Filetypes (rep,zip,mnt)

    then it should take next <FileSetfolder>

XML:-

<SourceFolder>
    <FileSetFolder>
      <Path>D:\apache-2.4.10\htdocs</Path>
      <FileType>rep</FileType>
      <FileType>zip</FileType>
      <FileType>mnt</FileType>
    </FileSetFolder>
    <FileSetFolder>
      <Path>D:\Download\ROSXcenterAutoArchive\ArchiveStorage\archive</Path>
      <FileType>mnt</FileType>
      <FileType>952230</FileType>
    </FileSetFolder>
  </SourceFolder>

答案1

使用XML小星（有时安装为xmlstarlet而不是xml）：

paths=( $( xml sel -t -v '//FileSetFolder/Path' file.xml ) )

for path in "${paths[@]}"; do
    filetypes=( $( xml sel -t -v "//FileSetFolder[Path=\"$path\"]/FileType" file.xml ) )

    for filetype in "${filetypes[@]}"; do
            printf 'Path "%s" has a filetype "%s"\n' "$path" "$filetype"
    done
done

输出：

Path "D:\apache-2.4.10\htdocs" has a filetype "rep"
Path "D:\apache-2.4.10\htdocs" has a filetype "zip"
Path "D:\apache-2.4.10\htdocs" has a filetype "mnt"
Path "D:\Download\ROSXcenterAutoArchive\ArchiveStorage\archive" has a filetype "mnt"
Path "D:\Download\ROSXcenterAutoArchive\ArchiveStorage\archive" has a filetype "952230"

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