在.NET中创建的OpenXML电子表格无法在iPad中打开

Question

我正在尝试在.NET中生成一个电子表格,当他不在办公室时,我的经理将在iPad上打开它.

电子表格在Windows PC上打开很好,但是当试图在iPad上打开时,它会显示"阅读文档时出错"(非常有用!)

通过使用OpenXML SDK Productivity工具上的"比较"功能和在iPad 上打开的文档,并通过在记事本中手动编辑错误文档的XML文件,我将其缩小到文件xl/_rels/workbook .xml.rels,用于存储工作簿中各部分的关系.

这是我用来生成WorkbookPart和引用的代码

    WorkbookPart workbookPart1 = document.AddWorkbookPart();

    WorkbookStylesPart workbookStylesPart1 = workbookPart1.AddNewPart<WorkbookStylesPart>("rId3");
    ThemePart themePart1 = workbookPart1.AddNewPart<ThemePart>("rId2");
    WorksheetPart worksheetPart1 = workbookPart1.AddNewPart<WorksheetPart>("rId1");

我的代码生成以下输出,该输出无法在iPad上打开.

      <?xml version="1.0" encoding="utf-8" ?> 
      <Relationships xmlns="http://schemas.openxmlformats.org/package/2006/relationships">
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/styles" Target="/xl/styles.xml" Id="rId3" /> 
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/theme" Target="/xl/theme/theme.xml" Id="rId2" /> 
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/worksheet" Target="/xl/worksheets/sheet.xml" Id="rId1" /> 
      </Relationships>

如果我更改Target属性的值以使用相对引用路径,给出以下输出,则它会在iPad上打开.

      <?xml version="1.0" encoding="utf-8" ?> 
      <Relationships xmlns="http://schemas.openxmlformats.org/package/2006/relationships">
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/styles" Target="styles.xml" Id="rId3" /> 
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/theme" Target="theme/theme.xml" Id="rId2" /> 
          <Relationship Type="http://schemas.openxmlformats.org/officeDocument/2006/relationships/worksheet" Target="worksheets/sheet.xml" Id="rId1" /> 
      </Relationships>

所以问题是:
如何更改我的.NET代码,以便输出具有相对路径的XML的第二个版本.

感谢所有的帮助!

Answer 1

我花了很多时间研究这个,并认为我会分享我的结果.看来OpenXML做了两件事.1. content_types.xml文件缺少工作簿2的条目.xl/_rels/workbook.xml.rels文件使用完全相对路径.

Excel本身打开文件很好,但我在iPad上尝试了各种应用程序,但都失败了.所以我不得不使用以下代码自己手动修复文件.它假定文件的整个内容作为流传递并使用DotNetZip打开和操作.希望这段代码能帮助别人!

    private Stream ApplyOpenXmlFix(Stream input)
    {
        const string RELS_FILE = @"xl/_rels/workbook.xml.rels";
        const string RELATIONSHIP_ELEMENT = "Relationship";
        const string CONTENT_TYPE_FILE = @"[Content_Types].xml";
        const string XL_WORKBOOK_XML = "/xl/workbook.xml";
        const string TARGET_ATTRIBUTE = "Target";
        const string SUPERFLUOUS_PATH = "/xl/";
        const string OVERRIDE_ELEMENT = "Override";
        const string PARTNAME_ATTRIBUTE = "PartName";
        const string CONTENTTYPE_ATTRIBUTE = "ContentType";
        const string CONTENTTYPE_VALUE = "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet.main+xml";

        XNamespace contentTypesNamespace = "http://schemas.openxmlformats.org/package/2006/content-types";
        XNamespace relsNamespace = "http://schemas.openxmlformats.org/package/2006/relationships";
        XDocument xlDocument;
        MemoryStream memWriter;

        try
        {
            input.Seek(0, SeekOrigin.Begin);
            ZipFile zip = ZipFile.Read(input);

            //First we fix the workbook relations file
            var workbookRelations = zip.Entries.Where(e => e.FileName == RELS_FILE).Single();
            xlDocument = XDocument.Load(workbookRelations.OpenReader());

            //Remove the /xl/ relative path from all target attributes
            foreach (var relationship in xlDocument.Root.Elements(relsNamespace + RELATIONSHIP_ELEMENT))
            {
                var target = relationship.Attribute(TARGET_ATTRIBUTE);

                if (target != null && target.Value.StartsWith(SUPERFLUOUS_PATH))
                {
                    target.Value = target.Value.Substring(SUPERFLUOUS_PATH.Length);
                }
            }

            //Replace the content in the source zip file
            memWriter = new MemoryStream();
            xlDocument.Save(memWriter, SaveOptions.DisableFormatting);
            memWriter.Seek(0, SeekOrigin.Begin);
            zip.UpdateEntry(RELS_FILE, memWriter);

            //Now we fix the content types XML file
            var contentTypeEntry = zip.Entries.Where(e => e.FileName == CONTENT_TYPE_FILE).Single();
            xlDocument = XDocument.Load(contentTypeEntry.OpenReader());

            if (!xlDocument.Root.Elements().Any(e =>
                e.Name == contentTypesNamespace + OVERRIDE_ELEMENT &&
                e.Attribute(PARTNAME_ATTRIBUTE) != null &&
                e.Attribute(PARTNAME_ATTRIBUTE).Value == XL_WORKBOOK_XML))
            {
                //Add in the missing element
                var overrideElement = new XElement(
                    contentTypesNamespace + OVERRIDE_ELEMENT,
                    new XAttribute(PARTNAME_ATTRIBUTE, XL_WORKBOOK_XML),
                    new XAttribute(CONTENTTYPE_ATTRIBUTE, CONTENTTYPE_VALUE));

                xlDocument.Root.Add(overrideElement);

                //Replace the content
                memWriter = new MemoryStream();
                xlDocument.Save(memWriter, SaveOptions.DisableFormatting);
                memWriter.Seek(0, SeekOrigin.Begin);
                zip.UpdateEntry(CONTENT_TYPE_FILE, memWriter);
            }

            Stream output = new MemoryStream();

            //Save file
            zip.Save(output);

            return output;
        }
        catch
        {
            //Just in case it fails, return the original document
            return input;
        }
    }

Answer 2

创建 OpenXML 电子表格后，您可以尝试验证它：

using System;
using System.Collections.Generic;
using DocumentFormat.OpenXml.Packaging;
using DocumentFormat.OpenXml.Validation;

using (OpenXmlPackage document = SpreadsheetDocument.Open(spreadsheetPathToValidate, false))
{
    var validator = new OpenXmlValidator();
    IEnumerable<ValidationErrorInfo> errors = validator.Validate(document);
    foreach (ValidationErrorInfo info in errors)
    {
        try
        {
            Console.WriteLine("Validation information: {0} {1} in {2} part (path {3}): {4}",
                        info.ErrorType,
                        info.Node.GetType().Name,
                        info.Part.Uri,
                        info.Path.XPath,
                        info.Description);
        }
        catch (Exception ex)
        {
            Console.WriteLine("Validation failed: {0}", ex);
        }
    }
}

希望有帮助，