我必须从不同的不同表中获取值，并且值不止一个，并且不同不同下拉框中的所有值都必须在框的基础上显示，我已经这样做了，不确定我的方法是否正确或不是，但我获得了值(value)，但最终当我尝试选择值时发生了什么，它即将到来，但像这样 [saurabh，prakash，karan] 并在带有数组符号的一行中，但我想要 saurabh 和下一行 prakash，所以这是我的问题，这里是我调用 empDetail 方法的 servlet 类
@WebServlet(name = "EmployeeMapping", urlPatterns = {"/EmployeeMapping"})
public class EmployeeMapping extends HttpServlet {
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
HttpSession session = request.getSession(true);
Employee emp1 = (Employee)session.getAttribute("emp1");
EmployeeBean eb = new EmployeeBean();
List<Employee> leave1 = eb.empDetail(emp1);
if(emp1.isValid())
{
request.setAttribute("leave1",leave1);
response.sendRedirect("AssignProject.jsp");
}}
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
@Override
public String getServletInfo() {
return "Short description";}}
这是我的方法
public List<Employee> empDetail(Employee emp1) {
List<Employee> leave1 = new ArrayList<Employee>();
List<String> myList1 = new ArrayList<String>();
List<String> myList2 = new ArrayList<String>();
List<String> myList3 = new ArrayList<String>();
List<String> myList4 = new ArrayList<String>();
try{
con = ConnectionManager.getConnection();
stmt1 = con.createStatement();
String Query11 = "SELECT projectname FROM addproject";
System.out.println("Query1 is" +Query11);
rs1 = stmt1.executeQuery(Query11);
while(rs1.next())
{
String z =(rs1.getString("projectname"));
System.out.println("...11." +z);
myList1.add(z);
}
emp1.setPrjname(myList1);
stmt2 = con.createStatement();
String Query12 = "select empName from empinfo where access_type ='teamleader'";
System.out.println("Query1 is" +Query12);
rs2 = stmt2.executeQuery(Query12);
while(rs2.next())
{
String p = rs2.getString("empName");
System.out.println("...11." +p);
myList2.add(p);
}
emp1.setTlname(myList2);
stmt3 = con.createStatement();
String Query13 = "select empName from empinfo where access_type ='manager'";
System.out.println("Query1 is" +Query13);
rs3 = stmt3.executeQuery(Query13);
while(rs3.next())
{
String q = rs3.getString("empName");
System.out.println("...11." +q);
myList3.add(q);
}
emp1.setManname(myList3);
stmt4 = con.createStatement();
String Query14 = "select empName from empinfo where access_type ='employee'";
System.out.println("Query1 is" +Query14);
rs4 = stmt4.executeQuery(Query14);
while(rs4.next())
{
String r = rs4.getString("empName");
System.out.println("...11." +r);
myList4.add(r);
}
emp1.setEmpname(myList4);
leave1.add(emp1);
emp1.setValid(true);
}
catch (SQLException ex) {
} finally {
try {
if (pstmt != null) {
pstmt.close();
}
if (con != null) {
con.close();
}
} catch (SQLException ex) {
Logger.getLogger(EmployeeBean.class.getName()).log(Level.SEVERE, null, ex);
}
}
return leave1;
};
这是我的jsp页面，其中我得到了像[saurabh, prakash, karan]这样的值，但是如果我给出${emp1.prjname}这个${emp1.prjname[0]}的insted，那么我得到的是saurabh所以明智地，我在这里面临着困难，不能用这个循环，所以请任何人帮助我..
<%if(session.getAttribute("emp1")==null)
{
String message ="please login first";
request.setAttribute("message", message);
request.getRequestDispatcher("index.jsp").forward(request, response);
}
%>
<title>RESOURCE MANAGEMENT SYSTEM</title>
<head
</head><body>
<%
Employee emp1 = (Employee)session.getAttribute("emp1");
session.setAttribute("emp1",emp1);
%>
<tr>
<td >select project name </td>
<td> <select name="pname">
<option value="${emp1.prjnamename}">${emp1.prjnamename}</option></select></td>
</tr>
<tr>
<td >manager </td>
<td> <select name="manager">
<option value="${emp1.manname}">${emp1.manname}</option></select></td>
</tr>
<tr>
<td >team leader</td>
<td> <select name="teamleader">
<option value="emp1.Tlname()">emp1.Tlname()</option></select></td>
</tr>
<tr>
<td> <section class="container">
<div>
<select id="leftValues" size="5" multiple>
<option value="${emp1.empname}">${emp1.empname}</option>
</select>
</div>
</td>
<td>
<div>
<input type="button" id="btnLeft" value="<<" />
<input type="button" id="btnRight" value=">>" />
</div>
</td>
<td>
<div>
<select name="nameofemps" id="rightValues" size="4" multiple="multiple">
<option>
</option>
</select>
<div>
<input type="text" id="txtRight" />
</div>
</div>
</section>
</td>
</tr>
</table><BR>
</fieldset>
</body>
</html>
最佳答案
您需要使用 request 属性迭代 forEach 循环，
<select id="leftValues" size="5" multiple>
<c:forEach var='temp' items='${leave1}' >
<option value="${temp.empname}">${temp.empname}</option>
</c:forEach>
</select>
此外，您正在此处创建一个新请求，
if(emp1.isValid())
{
request.setAttribute("leave1",leave1);
response.sendRedirect("AssignProject.jsp");
}}
相同的请求应该转发到jsp。使用 RequestDispatcher 代替
我假设您正在尝试打印请求中的leave1，如果您的观点不同，请在我的答案下方评论
关于java - 我在 jsp 页面上检索下拉选项的数组值时遇到问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26923974/