HtmlUnit Google 登录

Question

我想自动执行 google 登录并接受 oauth 以获取验证码。

我通过 nextButton.click() 得到一个 NullPointerExeption，我找不到 "next" 按钮。

        WebClient webClient = new WebClient(); //BrowserVersion.FIREFOX_38
        HtmlPage page = webClient.getPage("https://accounts.google.com/o/oauth2/auth?");            

        HtmlTextInput email = (HtmlTextInput)page.getElementById("Email");
        email.setValueAttribute(emailAddress);
        HtmlSubmitInput nextButton = (HtmlSubmitInput)page.getElementById("next");

        HtmlPage newPage = (HtmlPage)nextButton.click();
        webClient.waitForBackgroundJavaScriptStartingBefore(8000);

        HtmlTextInput passwd = (HtmlTextInput)page.getElementById("Passwd");
        passwd.setValueAttribute(password);
        HtmlSubmitInput signIn = (HtmlSubmitInput)page.getElementById("signIn");

        HtmlPage pageSucces = (HtmlPage)signIn.click();
        webClient.waitForBackgroundJavaScriptStartingBefore(8000);

        HtmlSubmitInput submitAccess = (HtmlSubmitInput)page.getElementById("submit_approve_access");
        HtmlPage pageAccess = (HtmlPage)submitAccess.click();
        webClient.waitForBackgroundJavaScriptStartingBefore(8000);

        HtmlTextInput code = (HtmlTextInput)page.getElementById("code");
        System.out.println(code.getText());

我试过了Can't log in to Google using HtmlUnit - Can't advance to web page for entering password 没有成功。

Answer 1

问题出在网站上，html 页面不像标准浏览器。登录页面如下所示：https://lh6.ggpht.com/v19i1LtG-IdZKSZ-rhtuflJJmVmZM3gd3uauQQyLFvJxXTxYi4t8ygCQXwutu1nq69mmna8=w351

您还可以查看页面对象，在那里您可以看到包含所有可用 ID 的 idMap。

        HtmlTextInput email = (HtmlTextInput)page.getElementById("Email");
        email.setValueAttribute(emailAddress);
        HtmlPasswordInput passwd = (HtmlPasswordInput)page.getElementById("Passwd");
        passwd.setValueAttribute(password);
        HtmlSubmitInput signInButton = (HtmlSubmitInput)page.getElementById("signIn");

        webClient.waitForBackgroundJavaScriptStartingBefore(8000);
        HtmlPage newPage = (HtmlPage)signInButton.click();

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