Question

编写以下代码以点击＆＃34;下一步按钮＆＃34;在HTML页面中打开新页面并获取每个页面的数据：

try (final WebClient webClient = new WebClient(BrowserVersion.CHROME)) {
        webClient.getOptions().setJavaScriptEnabled(true);
        java.util.logging.Logger.getLogger("com.gargoylesoftware").setLevel(Level.OFF);
        webClient.getOptions().setThrowExceptionOnFailingStatusCode(false);
        webClient.getOptions().setThrowExceptionOnScriptError(false);
        webClient.setAjaxController(new NicelyResynchronizingAjaxController());
        String url = "http://codal.ir/ReportList.aspx?1=&2=&3=&4=-1&5=%d8%af%d8%b9%d9%88%d8%aa&6=&7=&8=-1&9=-1&10=-1&11=&12=False&13=0";
        WebRequest webRequest = new WebRequest(new URL(url));
        webRequest.setCharset(StandardCharsets.UTF_8);
        HtmlPage page = webClient.getPage(url);
        for (int page_counter = 0; page_counter < number_of_codal_pages_for_going_back; page_counter++) {
            HtmlTable html_table = page.getHtmlElementById("ctl00_ContentPlaceHolder1_gvList");
            for (int i_h_table_row = html_table.getRowCount() - 1; i_h_table_row > 0; i_h_table_row--) {
                final HtmlTableRow html_row = html_table.getRow(i_h_table_row);
                String ticker = html_row.getCell(0).asText();
                System.out.println(ticker);
            }
            System.out.println("______________________________");
            HtmlForm form = page.getFormByName("aspnetForm");
            HtmlSubmitInput button = form.getInputByName("ctl00$ContentPlaceHolder1$ucPager1$btnNext");
            page = button.click();
        }
    }

但无论打印什么都重复（与第一页相同），似乎程序无法进入下一页。

为什么呢？

感谢。

Answer 1

这适用于最新的SNAPSHOT

        try (WebClient webClient = new WebClient(BrowserVersion.CHROME)) {
        String url = "http://codal.ir/ReportList.aspx?1=&2=&3=&4=-1&5=%d8%af%d8%b9%d9%88%d8%aa&6=&7=&8=-1&9=-1&10=-1&11=&12=False&13=0";

        HtmlPage page = webClient.getPage(url);
        System.out.println("______________________________");
        System.out.println(page.getElementById("ctl00_ContentPlaceHolder1_ucPager1_gridpager").asXml());
        System.out.println("______________________________");

        HtmlForm form = page.getFormByName("aspnetForm");
        HtmlSubmitInput button = form.getInputByName("ctl00$ContentPlaceHolder1$ucPager1$btnNext");
        page = button.click();

        System.out.println("______________________________");
        System.out.println(page.getElementById("ctl00_ContentPlaceHolder1_ucPager1_gridpager").asXml());
        System.out.println("______________________________");
    }

转到下一页没有工作（通过HtmlUnit）

1 个答案: