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Apache POIXML Exception: InvalidFormatException:Package should contain a content type part [M1.13]

Problem:
I received following Apache POIXML Exception while using “Apache POI(XSSF API)” for reading xlsx file”.
Exception occurred during playback of script [TC002_Login] [CRFCN0019E: RationalTestScriptException on line 45 of script TC002_Login – org.apache.poi.POIXMLException: org.apache.poi.openxml4j.exceptions.InvalidFormatException: Package should contain a content type part [M1.13].].
 
Below is the error displayed in log file:
  • exception_name = org.apache.poi.POIXMLException
  • exception_message = org.apache.poi.openxml4j.exceptions.InvalidFormatException: Package should contain a content type part [M1.13]
  • script_name = TC002_Login
  • script_id = TC002_Login.java
  • line_number = 45
exception_stack = at org.apache.poi.util.PackageHelper.open(PackageHelper.java:41)
at org.apache.poi.xssf.usermodel.XSSFWorkbook.(XSSFWorkbook.java:204)
at SuperHelperClass.SuperScriptHelperlibrary.updateExcelReportScriptname(SuperScriptHelperlibrary.java:1018)
at TC002_Login.testMain(TC002_Login.java:45)
at org.eclipse.equinox.internal.app.EclipseAppContainer.callMethodWithException(EclipseAppContainer.java:587)
at org.eclipse.equinox.internal.app.EclipseAppHandle.run(EclipseAppHandle.java:198)
at org.eclipse.core.runtime.internal.adaptor.EclipseAppLauncher.runApplication(EclipseAppLauncher.java:134)
at org.eclipse.core.runtime.internal.adaptor.EclipseAppLauncher.start(EclipseAppLauncher.java:104)
at org.eclipse.core.runtime.adaptor.EclipseStarter.run(EclipseStarter.java:380)
at org.eclipse.core.runtime.adaptor.EclipseStarter.run(EclipseStarter.java:235)
at org.eclipse.equinox.launcher.Main.invokeFramework(Main.java:648)
at org.eclipse.equinox.launcher.Main.basicRun(Main.java:603)
at org.eclipse.equinox.launcher.Main.run(Main.java:1465)
at org.eclipse.equinox.launcher.Main.main(Main.java:1438)
at org.eclipse.core.launcher.Main.main(Main.java:34)
at org.apache.poi.openxml4j.opc.ZipPackage.getPartsImpl(ZipPackage.java:178)
at org.apache.poi.openxml4j.opc.OPCPackage.getParts(OPCPackage.java:662)
at org.apache.poi.openxml4j.opc.OPCPackage.open(OPCPackage.java:269)
at org.apache.poi.util.PackageHelper.open(PackageHelper.java:39)
 
Line number 45 read:
updateExcelReportScriptname(DestReport); //DestReport is a string variable with path+filename
The function ‘updateExcelReportScriptname’ is defined as below:
public void updateExcelReportScriptname(String DestReport)
{
                        FileInputStream file;
 
                        try {
                                    file = new FileInputStream(DestReport);
                                    XSSFWorkbook workbook;
                                    try {
                                                workbook = new XSSFWorkbook(file); //Line NUmber 1018
                                                XSSFSheet mySheet1 = workbook.getSheetAt(0);
                                                XSSFRow myRow1 = null;
                                                XSSFCell myCell1 = null;
 
                                                XSSFCellStyle style = workbook.createCellStyle();
                                                style.setAlignment(CellStyle.ALIGN_JUSTIFY);
                                                style.setFillPattern(CellStyle.SOLID_FOREGROUND);
                                                style.setBorderBottom(CellStyle.BORDER_THIN);
                                                style.setBorderTop(CellStyle.BORDER_THIN);
                                                style.setBorderRight(CellStyle.BORDER_THIN);
                                                style.setBorderLeft(CellStyle.BORDER_THIN);
                                                style.setWrapText(true);
 
                                                String excelData1[][] = new String[500][500];
 
                                                int row = mySheet1.getPhysicalNumberOfRows();
 
                                                excelData1[row][0] = getScriptName();
                                                myRow1 = mySheet1.createRow(row);
                                                myCell1 = myRow1.createCell(0);
                                                myCell1.setCellValue(excelData1[row][0]);
                                                style.setFillForegroundColor(IndexedColors.LIGHT_YELLOW.getIndex());
                                                myCell1.setCellStyle(style);
 
                                                for(int cellNum = 1; cellNum <= 19; cellNum++)
                                                {
                                                            XSSFCell myCell = null; 
                                                            myCell = myRow1.createCell(cellNum);
                                                            myCell.setCellValue(“”);
                                                }
                                                file.close();       
                                                FileOutputStream outFile =new FileOutputStream(DestReport); 
                                                workbook.write(outFile);  
                                                outFile.close();
 
                                    } catch (IOException e) {
                                                // TODO Auto-generated catch block
                                                e.printStackTrace();
                                    }   
 
                        } catch (FileNotFoundException e) {
                                    // TODO Auto-generated catch block
                                    e.printStackTrace();
                        }
 
            }
Solution:
There are two reasons I could analyse which may cause this error:
1. When I try to read ‘.xls’ file with my code and for ‘.xlsx’ files. There are different classes for xls and xlsx files namely:
Classes to be used for XLS file: HSSFWorkbook & HSSFSheet
Classes to be used for XLSX file: XSSFSheet & XSSFSheet
Required one must be used accordingly.
2. Sometimes, the same error is thrown even when the correct file format is used. The reason for this could be any of the below:
  • File is password protected
  • File is corrupted. For me, this was the case and I realized it when I tried to open the file and got bellow error. 
Microsoft Excel Error for Corrupt File
 
 Hope this helps solve the problem. 

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