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Как запустить цикл в JavaFX по элементу

2014-02-07 3 views
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Я сделал 1 функцию в JavaFX Я новичок в java, не знаю много функций об этом, и, возможно, это глупый вопрос, но то, что я сделал, я сделал 1 функцию, которая делает HTTP-вызов Google и получить его ответ я нашел структуру результата и выбрать его, но его полная строка из 10 записей, я хочу, чтобы все 10 записей отдельно, например, запись являетсяКак запустить цикл в JavaFX по элементу

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    Ujjain, Madhya Pradesh, India&nbsp;-&nbsp;CTO at SITS SolutionS 
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    <span class="st">22 Oct 2013 <b>...</b> Hello, I'm working as a Chief Technical Officer in a well known company at Ujjain. <br /> I've 3 years experience in web development field.</span> 
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его 10 Li с классом г, но я хочу все это в sep. так что я могу показать это Ли по мне то, что я ищу это цикл в соответствии с классом элемента или всех этих элементов в 1 массив, так что я могу написать цикл для него

Коды:

private void callUrl(String keywords) throws Exception { 
     String url = "http://www.google.com/search?q="+URLEncoder.encode(keywords); 
     Document response = Jsoup.connect(url) 
          .userAgent("Mozilla/5.0 (Windows; U; WindowsNT 5.1; en-US; rv1.8.1.6) Gecko/20070725 Firefox/2.0.0.6") 
          .referrer("http://www.google.com").get(); 
     Elements records = response.select("body #ires li.g"); 
     //record = records.split(".g"); 
     System.out.println(records); System.exit(0); 
    }

источник

2014-02-07 Mohit Bumb

ответ

1

Понял сам

String url = "http://www.google.com/search?q="+URLEncoder.encode(keywords); 
     Document response = Jsoup.connect(url) 
          .userAgent("Mozilla/5.0 (Windows; U; WindowsNT 5.1; en-US; rv1.8.1.6) Gecko/20070725 Firefox/2.0.0.6") 
          .referrer("http://www.google.com").get(); 
     Elements records = response.select("body #ires li.g"); 
     for (int i = 0; i < records.size()-1; i++) { 
      System.out.println(records.get(i)); 
     }

источник

2014-02-08 06:52:41

1

Я все еще пытаясь понять полный ответ, и я сожалею, чтобы ответить на вопрос с другим вопросом.

Я использовал два шаблона, чтобы найти все места, где найдены <li> и </li>.

//Your string 
String bigString = "<div><li><div><li></li></div></li></div><li><div><li></li></div></li>"; 

Pattern li_start = Pattern.compile("<li>"); 
Pattern li_end = Pattern.compile("</li>"); 

Matcher m_li_start = li_start.matcher(bigString); 
Matcher m_li_end = li_end.matcher(bigString); 

List<Integer> list_of_li = new ArrayList<>(); 
List<Integer> list_of_li_close = new ArrayList<>(); 

    while (m_li_start.find()) { 
     int index_of_li = m_li_start.start(); //finds 1st index of <li> 
     list_of_li.add(index_of_li); 
    } 

    while (m_li_end.find()) {    
     int index_of_close_li = m_li_end.end(); //finds last index of </li> 
     list_of_li_close.add(index_of_close_li);    
    }

Когда я печатаю как список в консоли я получаю положение каждого <li> и </li>

[5, 14, 40, 49] //all index of <li> 
[13, 24, 42, 53] //all last index of </li>

Теперь мы знаем, что сочетание 1 пункта в каждом списке есть ответ, который вы ищете, в этом случае:

//print from [0,24] - 1st block of <li></li> 
//print from [29,53] - 2nd block of <li></li>

Теперь, как получить этот ответ до сих пор озадачивает меня, но я надеюсь, что это поможет вам по крайней мере немного. Снова, извините, я не мог помочь вам с лучшим ответом.

источник

2014-02-08 06:41:20 sandiego

+0

вау! ваш ответ показывает, почему я должен читать о записях перед рукой XD, так или иначе, я все еще новичок: P – sandiego

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