Question

我有这个动态页面，它根据数据统计变化。每当一页测试完成时，我都必须单击下一页。为此，我具有测试脚本，该脚本将从应用程序捕获页面计数。我在下面使用的JAVA代码不知道哪里出错了，因为我没有收到任何错误消息

// Page 1 is always present     
        Reporter.log("Successfully rendered View Portfolios Page 1");
    Add_Log.info("Successfully rendered View Portfolios Page 1");

//This is fetching page count from the application which I convert to Int
    String entriesTxt = driver.findElement(By.xpath("(//label[contains(@class,'pf-pageNo-label')])[2]")).getText().trim();
    String[] aEntriesText = entriesTxt.split(" ");
    String totalEntriesText = aEntriesText[aEntriesText.length - 1];
    int result = Integer.parseInt(totalEntriesText);    
    System.out.println(result);
            // Page 2
            // This if Page count is great than 1 then perform below action
    if (result > 1) {
    pagenew(driver, Filters[1]);
    Reporter.log("Successfully rendered View Portfolios Page 2");
    Add_Log.info("Successfully rendered View Portfolios Page 2");
    Thread.sleep(3000);
    }
    // Page 3
            // This if Page count is great than 2 then perform below action
    if (result < 2) {
    pagenew(driver, Filters[2]);
    Reporter.log("Successfully rendered View Portfolios Page 3");
    Add_Log.info("Successfully rendered View Portfolios Page 3");
    Thread.sleep(3000);
    }

使用上面的代码，它正在渲染此语句不起作用的所有页面如果（结果> 1）我希望如果result = 1

，则第2页不应该呈现

Answer 1

我最好的猜测是，而不是

...
if (result < 2) {
...

你想写

...
if (result > 2) {
...

因此，即使结果为1，也会呈现第3页。

Java如果值大于执行

1 个答案: