如何使用openxml sdk和c#在word文档中添加多级列表？

Question

我正在尝试使用 openxml sdk v2.5 和 C# 以编程方式在 word 文档中实现多级列表。我使用以下代码在多级列表中显示标题。我正在调用 AddHeading 方法并传递相应的参数，如下所示。我的预期结果如下。

1.  Parent
    1.1. childItem
      1.1.1. subchildItem
    1.2. childItem
2.  Parent
    2.1. childItem
3.  Parent

但我得到的输出是

1. Parent
    1. childItem
      1. subchildItem
    2. childItem
2.  Parent
    1. childItem
3.  Parent

public static void AddHeading(WordprocessingDocument document, string colorVal, int fontSizeVal, string styleId, string styleName, string headingText, int numLvlRef, int numIdVal)
        {
            StyleRunProperties styleRunProperties = new StyleRunProperties();
            Color color = new Color() { Val = colorVal };
            DocumentFormat.OpenXml.Wordprocessing.FontSize fontSize1 = new DocumentFormat.OpenXml.Wordprocessing.FontSize();
            fontSize1.Val = new StringValue(fontSizeVal.ToString());

            styleRunProperties.Append(color);
            styleRunProperties.Append(fontSize1);
            AddStyleToDoc(document.MainDocumentPart.Document.MainDocumentPart, styleId, styleName, styleRunProperties, document);

            Paragraph p = new Paragraph();
            ParagraphProperties pp = new ParagraphProperties();
            pp.ParagraphStyleId = new ParagraphStyleId() { Val = styleId };
            pp.SpacingBetweenLines = new SpacingBetweenLines() { After = "0" };

            ParagraphStyleId paragraphStyleId1 = new ParagraphStyleId() { Val = "ListParagraph" };
            NumberingProperties numberingProperties1 = new NumberingProperties();
            NumberingLevelReference numberingLevelReference1 = new NumberingLevelReference() { Val = numLvlRef };

            NumberingId numberingId1 = new NumberingId() { Val = numIdVal }; //Val is 1, 2, 3 etc based on your numberingid in your numbering element

            numberingProperties1.Append(numberingLevelReference1); Indentation indentation1 = new Indentation() { FirstLineChars = 0 };
            numberingProperties1.Append(numberingId1);
            pp.Append(paragraphStyleId1);
            pp.Append(numberingProperties1);
            pp.Append(indentation1);
            p.Append(pp);
            Run r = new Run();
            Text t = new Text(headingText) { Space = SpaceProcessingModeValues.Preserve };
            r.Append(t);
            p.Append(r);

            document.MainDocumentPart.Document.Body.Append(p);
        }


      public static void AddStyleToDoc(MainDocumentPart mainPart, string styleid, string stylename, StyleRunProperties styleRunProperties, WordprocessingDocument document)
        {
            StyleDefinitionsPart part = mainPart.StyleDefinitionsPart;
            if (part == null)
            {
                part = AddStylesPartToPackage(mainPart);
                AddNewStyle(part, styleid, stylename, styleRunProperties);
            }
            else
            {
                if (IsStyleIdInDocument(mainPart, styleid) != true)
                {
                    string styleidFromName = GetStyleIdFromStyleName(document, stylename);
                    if (styleidFromName == null)
                    {
                        AddNewStyle(part, styleid, stylename, styleRunProperties);
                    }
                    else
                        styleid = styleidFromName;
                }
            }
        }

        public static string GetStyleIdFromStyleName(WordprocessingDocument doc, string styleName)
        {
            StyleDefinitionsPart stylePart = doc.MainDocumentPart.StyleDefinitionsPart;
            string styleId = stylePart.Styles.Descendants<StyleName>()
                .Where(s => s.Val.Value.Equals(styleName) &&
                    (((Style)s.Parent).Type == StyleValues.Paragraph))
                .Select(n => ((Style)n.Parent).StyleId).FirstOrDefault();

            return styleId;
        }

        public static StyleDefinitionsPart AddStylesPartToPackage(MainDocumentPart mainPart)
        {
            StyleDefinitionsPart part;
            part = mainPart.AddNewPart<StyleDefinitionsPart>();
            DocumentFormat.OpenXml.Wordprocessing.Styles root = new DocumentFormat.OpenXml.Wordprocessing.Styles();
            root.Save(part);
            return part;
        }

        public static bool IsStyleIdInDocument(MainDocumentPart mainPart, string styleid)
        {
            DocumentFormat.OpenXml.Wordprocessing.Styles s = mainPart.StyleDefinitionsPart.Styles;

            int n = s.Elements<DocumentFormat.OpenXml.Wordprocessing.Style>().Count();
            if (n == 0)
                return false;

            DocumentFormat.OpenXml.Wordprocessing.Style style = s.Elements<DocumentFormat.OpenXml.Wordprocessing.Style>()
                .Where(st => (st.StyleId == styleid) && (st.Type == StyleValues.Paragraph))
                .FirstOrDefault();
            if (style == null)
                return false;

            return true;
        }

        private static void AddNewStyle(StyleDefinitionsPart styleDefinitionsPart, string styleid, string stylename, StyleRunProperties styleRunProperties)
        {
            DocumentFormat.OpenXml.Wordprocessing.Styles styles = styleDefinitionsPart.Styles;

            DocumentFormat.OpenXml.Wordprocessing.Style style = new DocumentFormat.OpenXml.Wordprocessing.Style()
            {
                Type = StyleValues.Paragraph,
                StyleId = styleid,
                CustomStyle = false
            };
            style.Append(new StyleName() { Val = stylename });
            style.Append(new BasedOn() { Val = "Normal" });
            style.Append(new NextParagraphStyle() { Val = "Normal" });
            style.Append(new UIPriority() { Val = 900 });

            styles.Append(style);
        }

Answer 1

要在 Microsoft Word 中使用多级列表，您需要确保：

1. 您已在编号定义部分正确设置了所需的多级列表（这意味着您的 w:numbering 元素包含指定不同列表级别的 w:num 和对应的 w:abstractNum）；
2. 每个列表级别都有一种样式，它引用w:num 元素指定的编号ID 和所需的列表级别；和
3. 您的段落引用了列表级别的正确样式。

请查看我对如何create multi-level ordered lists with Open XML 和display multi-level lists in Word documents using C# and the Open XML SDK 的回答以获取更多详细信息和解释。