线程“main”java.lang.NullPointerException与.xlsx文件中出现错误异常

我有一张表格，它包括：号码、姓名、通行证、通行证、结果。我想要获取testweb的名称、传递、结果，我尝试编写它，但出现以下错误：

String user[],pass[],result[];
Workbook workbook = WorkbookFactory.create(new File(pathFile));
Sheet sheet = workbook.getSheetAt(0);
workbook.close();
if (sheet != null) {
int rowindex =0;
for(Row row : sheet) {
   for(Cell cell:row){
      int i=cell.getColumnIndex();
      if(i==1) user[rowindex]=cell.toString();
      if(i==2) pass[rowindex]=cell.toString();
      if(i==4) result[rowindex]=cell.toString();
      }
      rowindex++;
  }
} else {
    System.out.println("Sheet was not found");
  }    

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在你的问题中没有太多的细节，但根据你的代码，我猜你会这样做

 FileInputStream inputStream = new FileInputStream(new File(file));
 Workbook workbook = new XSSFWorkbook(inputStream);
 Sheet sheet = workbook.getSheetAt(0);

 if (sheet != null) {
    int n = sheet.getLastRowNum();
    // Do the rest of you code
} else {
    System.out.println("Sheet was not found Check to make sure file was read and sheet was access properly.");
}          

我猜你的工作表是空的。因此，您需要在检查中输入一些保护代码，以确保您拥有有效的工作表

像这样的

 FileInputStream inputStream = new FileInputStream(new File(file));
 Workbook workbook = new XSSFWorkbook(inputStream);
 Sheet sheet = workbook.getSheetAt(0);

 if (sheet != null) {
    int n = sheet.getLastRowNum();
    // Do the rest of you code
} else {
    System.out.println("Sheet was not found Check to make sure file was read and sheet was access properly.");
}          

它起作用了

   if (sheet != null) {  
          rowNum=sheet.getLastRowNum()+1; //getlastrow() return index last row
          String username[]=new String[rowNum];
          int index=0;
          Iterator<Row> rowIterator = sheet.iterator();
                while (rowIterator.hasNext()) {
                        Row row = rowIterator.next();
                        Iterator <Cell> cellIterator = row.cellIterator();
                        index++;
                           while (cellIterator.hasNext()) {
                               Cell cell = cellIterator.next();
                               if(cell.getColumnIndex()==1) 
                               username[index]=cell.toString();
                            }
                   }

     } else {
        System.out.println(" Sheet was not found ");
     }

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请提供所有涉及的代码和堆栈跟踪。因此，如果您使用的是ApachePOI，那么您应该使用XSSFWorkbook