如何使用Selenium WebDriver和Java查找损坏的链接

问题描述

我想验证网站上已损坏的链接，我正在使用此代码：

I want to verify broken links on a website and I am using this code:

 public static int invalidLink;
    String currentLink;
    String temp;

    public static void main(String[] args) throws IOException {
        // Launch The Browser
        WebDriver driver = new FirefoxDriver();
        // Enter URL
        driver.get("http://www.applicoinc.com");

        // Get all the links URL
        List<WebElement> ele = driver.findElements(By.tagName("a"));
        System.out.println("size:" + ele.size());
        boolean isValid = false;
        for (int i = 0; i < ele.size(); i++) {

            isValid = getResponseCode(ele.get(i).getAttribute("href"));
            if (isValid) {
                System.out.println("ValidLinks:" + ele.get(i).getAttribute("href"));
                driver.get(ele.get(i).getAttribute("href"));
                List<WebElement> ele1 = driver.findElements(By.tagName("a"));
                System.out.println("InsideSize:" + ele1.size());
                for (int j=0; j<ele1.size(); j++){
                    isValid = getResponseCode(ele.get(j).getAttribute("href"));
                    if (isValid) {
                        System.out.println("ValidLinks:" + ele.get(j).getAttribute("href"));
                    }
                    else{
                        System.out.println("InvalidLinks:"+ ele.get(j).getAttribute("href"));
                    }
                }

                } else {
                    System.out.println("InvalidLinks:"
                            + ele.get(i).getAttribute("href"));
                }

            }
        }
    }


    public static boolean getResponseCode(String urlString) {
        boolean isValid = false;
        try {
            URL u = new URL(urlString);
            HttpURLConnection h = (HttpURLConnection) u.openConnection();
            h.setRequestMethod("GET");
            h.connect();
            System.out.println(h.getResponseCode());
            if (h.getResponseCode() != 404) {
                isValid = true;
            }
        } catch (Exception e) {

        }
        return isValid;
    }

}

推荐答案

我会保持它返回一个int，只是让MalformedURLException成为一个特例，返回-1。

I would keep it returning an int, and just have the MalformedURLException be a special case, returning -1.

public static int getResponseCode(String urlString) {
    try {
        URL u = new URL(urlString);
        HttpURLConnection h =  (HttpURLConnection)  u.openConnection();
        h.setRequestMethod("GET");
        h.connect();
        return h.getResponseCode();

    } catch (MalformedURLException e) {
        return -1;
    }
}

编辑：看来你坚持使用布尔值方法，正如我之前所说的那样有它的局限性，但是应该可以用于演示目的。

It seems you're sticking with the boolean approach, as I said before this has it's limitations but should work ok for demonstartion purposes.

没有理由第二次找到你所采用的方法来找到所有元素。试试这个：

There is no reason to find all elements a second time taking the approach you have. Try this:

// Get all the links
List<WebElement> ele = driver.findElements(By.tagName("a"));
System.out.println("size:" + ele.size());
boolean isValid = false;
for (int i = 0; i < ele.size(); i++) {
    string nextHref = ele.get(i).getAttribute("href");
    isValid = getResponseCode(nextHref);
    if (isValid) {
        System.out.println("Valid Link:" + nextHref);

    }
    else {
        System.out.println("INVALID Link:" + nextHref);

    }
}

这是未经测试的代码，所以如果它不起作用，请提供更多细节，而不仅仅是说'它不起作用'，提供输出和放大器。任何堆栈跟踪/错误消息（如果可能）。干杯

This is untested code, so if it does not work, please provide more detail than just saying 'it doesn't work', provide output & any stack traces/error messages if possible. Cheers

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