Question

我是新玩框架工作，我发现它有点困难。我正在从数据库中检索客户端名称列表并将其填充到下拉列表中，这是我的client.java代码

   package models;

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.util.*;

import play.db.ebean.Model;
public class Client extends Model {

    /**
     * 
     */
    private static final long serialVersionUID = -1932214701504374792L;
    public static String ClientName;
    public static ArrayList<String> Clientdetail= new ArrayList<>();
    public static ArrayList<String> PopulateClient() {
        try {
            Class.forName("com.microsoft.sqlserver.jdbc.SQLServerDriver").newInstance();
            Connection conn = DriverManager.getConnection("jdbc:sqlserver://localhost:1433","SUMEET","sumeet");
            Statement sta = conn.createStatement();
            String Sql = "select * from client";
            ResultSet rs = sta.executeQuery(Sql);
            while (rs.next()) {
                ClientName = rs.getString("ClientName");
                Clientdetail.add(ClientName);
               }

        } catch (InstantiationException | IllegalAccessException
                | ClassNotFoundException |SQLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return(Clientdetail);

    }

}

这是我的application.java代码

package controllers;

import models.Client;

import play.mvc.*;
import views.html.*;

public class Application extends Controller {


    public static Result index(){

        return ok(index.render(Client.PopulateClient()));
    }

}

这是我的index.scala.html

    @(ClientDetails: java.util.ArrayList[String])

@main("ADMS") {

   <center>
    <form id="select">
   <a>CONSULTANT</a>
       <select name=Consultant>
           <option value="lokesh">Lokesh</option>
           <option>@ClientDetails</option>
           <option>Vidyasekar</option>
           <option>Abhishek</option>
           <option>Naveen</option>
           <option>Nanda</option>
       </select>
     <table border="1">
       <tr>
       <td width=50px>Client</td>
       <td width=50px>Project</td>
       <td width=50px>Task</td>
       <td width=50px>Date</td>
       <td width=50px>Consultant</td>
       <td width=50px>Role</td>
       <td width=80px>Is Billable</td>
       </tr>
       <tr>
       <td>@ClientDetails</td>
       </tr>
       </table>
      </form>
   </center>
}

main.scala.html

@(title: String)(Content: Html)


<!DOCTYPE html>

<html>
    <head>
        <title>@title</title>
   </head>
    <body>
    @Content
    </body>
</html>

有人可以帮助我吗？我需要使用数组值填充下拉列表，并且填充的数据只是括号 - ＆gt; “[]”

Answer 1

Play框架提供模板帮助程序库，它提供了使用选项和选定值构建选择下拉列表的功能。一旦理解得当，使用起来非常简单。

视图中的@helper.select（）方法采用与select类型的输入字段相关的各种参数。第一个参数是表单字段，因为在这里我们没有任何形式我们可以创建一个临时表单并在其中创建一个名为Consultant的新字段[因为这将是select字段的name属性的值]。第二个参数是选项的映射，其中键和值分别对应于选项标签的值和选项标签中包含的文本。

控制器代码

package controllers;

import models.Client;

import play.mvc.*;
import views.html.*;

public class Application extends Controller {


    public static Result index(){

        return ok(index.render(Client.getClientDetails()));
    }

}

型号代码

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Statement;
import java.util.*;

import play.db.ebean.Model;
public class Client extends Model {

    /**
     * 
     */
    private static final long serialVersionUID = -1932214701504374792L;
    public static String ClientName;
    public static HashMap<String, String> Clientdetail= new HashMap<String, String>();
    public static HashMap<String, String> getClientDetails() {
        try {
            Class.forName("com.microsoft.sqlserver.jdbc.SQLServerDriver").newInstance();
            Connection conn = DriverManager.getConnection("jdbc:sqlserver://localhost:1433","SUMEET","sumeet");
            Statement sta = conn.createStatement();
            String Sql = "select * from client";
            ResultSet rs = sta.executeQuery(Sql);
            while (rs.next()) {
                ClientName = rs.getString("ClientName");
                Clientdetail.put(ClientName,ClientName);
               }

        } catch (InstantiationException | IllegalAccessException
                | ClassNotFoundException |SQLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        return(Clientdetail);

    }

}

查看代码：index.scala.html

@(ClientDetails: java.util.HashMap[String, String])

@import helper._

@main("ADMS") {

   <center>
    <form id="select">
   <a>CONSULTANT</a>
   @select(Form.form()("Consultant"),           
       options(ClientDetails),
       'value -> "clientName1"[any value that should be selected by default])
     <table border="1">
       <tr>
       <td width=50px>Client</td>
       <td width=50px>Project</td>
       <td width=50px>Task</td>
       <td width=50px>Date</td>
       <td width=50px>Consultant</td>
       <td width=50px>Role</td>
       <td width=80px>Is Billable</td>
       </tr>
       <tr>
       <td>@ClientDetails</td>
       </tr>
       </table>
      </form>
   </center>
}

Answer 2

是的，不幸的是Play的模板系统是基于Scala的，所以你必须映射你的@ClientDetails：

@ClientDetails.map { clientName =>
  <option>@clientName</option>
}

如果你返回更复杂的对象，它可能看起来更像这样：

@ClientDetails.map { client =>
  <option value="@client.id">@client.name</option>
}

我希望这有帮助！

使用播放帧工作中的数据库中的值填充下拉列表

2 个答案: