JAXB从XML内的文件路径读取XML1个主XSD和1个子XSD

编程问答 2022-06-10

问题描述

请帮助建议通过JAXB读取xml内部xmls文件路径的最佳方法。实施需要提供最佳性能。我们在下面尝试过

Master XML Example生成JAXB类

    <Header RefNum="SIM0000053728SIM" BU="CSGBU" Format="xml" xml:lang="en" RetStat="" Ind="0" Type="INV" Postprocmethod="L"/>
    <Part File="file:C:\file\xml\SIM0000053728SIM\ADDR1710135.88361.1.1.0.xml" BillSeqNo="88361" LinkType="ABS" DocType="ADD" Format="XML"/>
    <Part File="file:C:\file\xml\SIM0000053728SIM\AGGCHG1710135.88361.1.1.0.xml" BillSeqNo="88361" LinkType="ABS" DocType="AGG" Format="XML"/>
    <Part File="file:C:\file\xml\SIM0000053728SIM\BAL1710135.88361.1.1.0.xml" BillSeqNo="88361" LinkType="ABS" DocType="BAL" Format="XML"/>

每个包含的文件RootElement创建JAXB类并进行迭代

   BusinessDocument finalObject =BusinessDocument) unMarshall.unmarshal(new File(colllectionXMLFile));
    List<Part> invoiceXMLList  = finalObject.getPart();
    for(Part part:invoiceXMLList) {
                    File file = new File(new URI(part.getFile()));
                    Document document  = (Document) unMarshall.unmarshal(file);
                    businessDocument.getDocument().add(document);
                }

我们的问题

1. Is there any other better approach with 1 XSD and 1 Unmarshall to improve performance
2. Is there any way to Merge all XMLs before Unmarshall to improve performance.

解决方法

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javajavajaxbjaxbunmarshallingxmlxmlxmlxmlxmlxml