Java和Apache-poi打开工作簿会引发exception

这是我的testing用例：

import java.io.File; import java.io.IOException; import org.apache.poi.openxml4j.exceptions.InvalidFormatException; import org.apache.poi.ss.usermodel.Workbook; import org.apache.poi.ss.usermodel.WorkbookFactory; import org.junit.Before; import org.junit.Test; public class TestWB { File sourceA; File sourceB; @Before public void setUp() throws Exception { sourceA = new File("C:\\junit\\sourceA.xls"); sourceB = new File("C:\\junit\\sourceB.xls"); } @Test public final void testValid() { try { Workbook wb = WorkbookFactory.create(sourceA); } catch (InvalidFormatException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } } @Test public final void testInValid() { try { Workbook wb = WorkbookFactory.create(sourceB); } catch (InvalidFormatException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } } } 

SourceA.xls传递和加载的地方，SourceB.xls是麻烦的。 这里是例外代码：

 java.lang.RuntimeException: Unexpected record type (org.apache.poi.hssf.record.StringRecord) at org.apache.poi.hssf.record.aggregates.RowRecordsAggregate.<init>(RowRecordsAggregate.java:97) at org.apache.poi.hssf.model.InternalSheet.<init>(InternalSheet.java:172) at org.apache.poi.hssf.model.InternalSheet.createSheet(InternalSheet.java:127) at org.apache.poi.hssf.usermodel.HSSFWorkbook.<init>(HSSFWorkbook.java:291) at org.apache.poi.ss.usermodel.WorkbookFactory.create(WorkbookFactory.java:88) at TestWB.testInValid(TestWB.java:35) ... 

如果我打开xls – 根据我所知，改变任何东西 – 保存它。 它的工作原理应该如此。 我怎样才能解决这个问题，而无需用户手动操作（打开 – 保存 – closures）

谢谢