小编Ada*_*ink的帖子

如何在Intelllij 13中运行playframework示例?

我正在使用Intellij 13 Ultimate并希望创建一个Play Framework示例.但我无法构建此项目,因为它总是抛出此错误:

object index is not a member of package views.html
Ok(views.html.index("Your new application is ready."))
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我在Mac和Windows平台上都试过这个,但错误总是一样的.

我怎么解决这个问题?

scala intellij-idea playframework

5
推荐指数
1
解决办法
2735
查看次数

如何从java或Android中的字符串生成XML文档?

我试图从Java创建一个动态生成的XML文件.这是我尝试使用的代码:

try{
            BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));

            System.out.println("how many elements: ");
            String str = bf.readLine();
            int no = Integer.parseInt(str);

            System.out.println("enetr root: ");
            String root = bf.readLine();

            DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
            DocumentBuilder db = dbf.newDocumentBuilder();
            Document d1 = db.newDocument();
            Element e1 = d1.createElement(root);
            d1.appendChild(e1);

            for (int i = 0; i < no; i++) {
                System.out.println("enter element: ");
                String element = bf.readLine();

                System.out.println("enter data: ");
                String data = bf.readLine();
                Element em = d1.createElement(element);
                em.appendChild(d1.createTextNode(data));

                e1.appendChild(em);
            }
            TransformerFactory tf = TransformerFactory.newInstance();
            Transformer …
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java xml android file

4
推荐指数
2
解决办法
1万
查看次数

要检查的RSpec测试(Rails模型:验证:某事,:includes =>来自其他activeRecord的列表)

以下是我使用的代码:

这是我的app/models档案:

class One < ActiveRecord::Base
  attr_accessible :name, :surname      
end

class Two < ActiveRecord::Base  
  attr_accessible :name, :one_surname

  # generate a list of one's surnames
  def self.LIST_SURNAMES
    list  = Array.new
    arr   = One.all
    arr.each {|one| list << one.surname}

    return list
  end

  validates :one_surname, :inclusion => self.LIST_SURNAMES()
end
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这是我的/spec/models档案:

FactoryGirl.define do
  factory :two do
    name "two_name"
  end

  factory :one do
    name "one_name"
    surname "one_surname"
  end    
end

describe Two do
  it 'should be created' do
    @one = FactoryGirl.create(:one)

    puts …
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rspec ruby-on-rails rspec2 factory-bot

3
推荐指数
1
解决办法
1078
查看次数

cakePHP 2.0 find('threaded')问题

我在我的Jobs控制器中运行以下命令.

$this->set('jobs', $this->Job->find('threaded', array('conditions' => array('Job.id' => 20))));
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现在在我看来我$jobsforeach循环中显示很好,但我的问题是我有使用字段parent_id链接到Job.id的子项.我知道链接工作正常,因为我可以看到数组中的子节点.

    Array
(
[0] => Array
    (
        [Job] => Array
            (
                [id] => 20
                [parent_id] => 0
                [rght] => 2
                [lft] => 1
                [client_id] => tasd
                [contact] => asdf
                [email] => sdf
                [address] => 
                [lat] => 
                [long] => 
                [user_id] => 1
                [request_type_id] => Electrical
                [date_start] => 0000-00-00 00:00:00
                [date_end] => 0000-00-00 00:00:00
                [date_complete] => 0000-00-00 00:00:00
                [date_closed] => 0000-00-00 00:00:00
                [status] => completed
                [brief_desc] => aasdf
                [desc] => …
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php cakephp

2
推荐指数
1
解决办法
4323
查看次数

添加类并在div之前插入

我希望能够将类添加"wp"到我"sub-menu dropdown-menu"班级之前的每个锚标记中.我该怎么做呢?

我目前的HTML:

  <a href="google.com">one</a>

  <ul class="sub-menu dropdown-menu">
   <li id="menu-item-5">
    <a href="">five</a>
  </li>
  </ul>

  <a href="google.com">one</a>
  <ul class="sub-menu dropdown-menu">
    <li id="menu-item-5">
      <a href="">five</a>
    </li>
  </ul>
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期望的输出

 <a href="google.com" class="wp">one</a>

 <a href="google.com" class="wp">one</a>
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jquery addclass insert

1
推荐指数
1
解决办法
1562
查看次数

java.lang.IllegalStateException:指定的子级已有父级.您必须首先在子项的父项上调用removeView()[对于表]

table在我的代码中动态绘制.在tablerow我的布局,我只是在XML硬编码得到的.我不知道为什么我会收到这个错误,完全没有想法.

 TableRow tr = (TableRow) findViewById(R.id.tr);
        TableRow.LayoutParams rowParams = new TableRow.LayoutParams(TableRow.LayoutParams.MATCH_PARENT, TableRow.LayoutParams.WRAP_CONTENT);

    TextView textView1 = new TextView(this);
    textView1.setGravity(Gravity.CENTER);
    textView1.setBackgroundResource(R.drawable.cell_shape);
    textView1.setText("1");
    textView1.setLayoutParams(rowParams);
    tr.addView(textView1);

    TextView textView2 = new TextView(this);
    textView2.setGravity(Gravity.CENTER);
    textView2.setBackgroundResource(R.drawable.cell_shape);
    textView2.setText("2");
    textView2.setLayoutParams(rowParams);
    tr.addView(textView2);

    TextView textView3 = new TextView(this);
    textView3.setGravity(Gravity.CENTER);
    textView3.setBackgroundResource(R.drawable.cell_shape);
    textView3.setText("3");
    textView3.setLayoutParams(rowParams);
    tr.addView(textView3);

    TextView textView4 = new TextView(this);
    textView4.setGravity(Gravity.CENTER);
    textView4.setBackgroundResource(R.drawable.cell_shape);
    textView4.setText("4");
    textView4.setLayoutParams(rowParams);
    tr.addView(textView4);

    TextView textView5 = new TextView(this);
    textView5.setGravity(Gravity.CENTER);
    textView5.setBackgroundResource(R.drawable.cell_shape);
    textView5.setText("5");
    textView5.setLayoutParams(rowParams);
    tr.addView(textView5);

    TextView textView6 = new TextView(this);
    textView6.setGravity(Gravity.CENTER);
    textView6.setBackgroundResource(R.drawable.cell_shape);
    textView6.setText("6");
    textView6.setLayoutParams(rowParams);
    tr.addView(textView6);

    TextView textView7 = (TextView) …
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android tablerow android-tablelayout

1
推荐指数
1
解决办法
8526
查看次数