Apache POI，抛出 UnsupportedZipFeatureException

Question

我正在使用 Apache POI 来读取 xsl 文件。我的pom文件更新：

    <dependency>
        <groupId>org.apache.poi</groupId>
        <artifactId>poi</artifactId>
        <version>5.2.0</version>
    </dependency>
    <dependency>
        <groupId>org.apache.poi</groupId>
        <artifactId>poi-ooxml</artifactId>
        <version>5.2.0</version>
    </dependency>

我的代码：

  public List<String> getSheetNames(String fileLocation) throws Exception {
    FileInputStream file = new FileInputStream(new File(fileLocation));
    Workbook workbook = null;
    try {
        workbook = new XSSFWorkbook(file);
    } catch (Exception e) {
        e.printStackTrace();
    }
    //do something with read lines
  }

这对于某些文件来说效果很好。但是，对于其他人来说，它会抛出此错误：

org.apache.commons.compress.archivers.zip.UnsupportedZipFeatureException: Unsupported feature compression method used in entry ...
... 
at org.apache.commons.compress.archivers.zip.ZipUtil.checkRequestedFeatures(ZipUtil.java:353)
    at org.apache.commons.compress.archivers.zip.ZipArchiveInputStream.read(ZipArchiveInputStream.java:514)
    at java.io.FilterInputStream.read(FilterInputStream.java:133)
    at org.apache.poi.openxml4j.util.ZipArchiveThresholdInputStream.read(ZipArchiveThresholdInputStream.java:80)
    at org.apache.poi.util.IOUtils.toByteArray(IOUtils.java:185)
    at org.apache.poi.util.IOUtils.toByteArray(IOUtils.java:153)
    at org.apache.poi.util.IOUtils.toByteArray(IOUtils.java:140)
    at org.apache.poi.openxml4j.util.ZipArchiveFakeEntry.<init>(ZipArchiveFakeEntry.java:72)
    at org.apache.poi.openxml4j.util.ZipInputStreamZipEntrySource.<init>(ZipInputStreamZipEntrySource.java:98)
    at org.apache.poi.openxml4j.opc.ZipPackage.<init>(ZipPackage.java:132)
    at org.apache.poi.openxml4j.opc.OPCPackage.open(OPCPackage.java:312)
    at org.apache.poi.ooxml.util.PackageHelper.open(PackageHelper.java:59)
    at org.apache.poi.xssf.usermodel.XSSFWorkbook.<init>(XSSFWorkbook.java:304)
    at org.apache.poi.xssf.usermodel.XSSFWorkbook.<init>(XSSFWorkbook.java:300)

此错误的修复方法是什么？我在网上找不到太多帮助。

更新：这现在不适用于任何文件。

Answer 1

像这样进行更改。如果文件是 xls 类型，则必须使用 HSSFWorkbook obj 来读取它

public List<String> getSheetNames(String fileLocation) throws Exception{
    File inputFile = new File(fileLocation);
    FileInputStream fis = new FileInputStream(inputFile);
    Workbook workbook = null;
            
    try {
        if (inputFile.getName().endsWith(".xlsx")) {
            workbook = new XSSFWorkbook(fis);
        } else if (inputFile.getName().endsWith(".xls")) {
            workbook = new HSSFWorkbook(fis);
        } else{
            fis.close();
            throw new IOException("This file extension type is not supported!");
        }
    } catch (Exception e) {
        e.printStackTrace();
    }
    //do something with read lines
}