在获取数据时动态传递FileName?硒-java

时间:2018-09-21 02:21:37

标签: java selenium selenium-webdriver automated-tests apache-poi

当我这样传递文件名时,它不会获取数据..如何在方法中动态传递fileName呢?

    public static String fetchData(int index,int rowNum, int cellNum,String FileName) throws FileNotFoundException {

        File filePath = new File(System.getProperty("user.dir") + "/InEdge_Files/ExcelData/" + FileName + ".xlsx");
        fis = new FileInputStream(filePath);
        sheet = workbook.getSheetAt(index);
        value = sheet.getRow(rowNum).getCell(cellNum).getStringCellValue();
        return value;

    }

...传递我使用的数据时,driver.get(fetchData(0,1,2,“ Data”);

1 个答案:

答案 0 :(得分:0)

String urlToOpen = fetchData(0,1,2,"Data");
driver.navigate().to(urlToOpen);