JSP连接SQL做select下拉菜单时，value与option的值相同时，总是显示option外的值有误，请帮我改下吧！谢谢~~
- Java code
<select name="name"> <% Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); Connection con = DriverManager.getConnection("jdbc:odbc:tscgdb","sa","sa"); Statement stat = con.createStatement(); ResultSet rs = stat.executeQuery("select * from books"); while(rs.next()){ %> <option value="<%=rs.getString("bookname")%>"> <%=rs.getString("bookname")%></option> <% } rs.close(); %> </select>系统现实“<%=rs.getString("bookname")%></option>”有误
------解决方案--------------------------------------------------------
你可以把rs.getString("bookname")得到的值放到一个临时变量中，再分别把它赋给value值和显示值。
- HTML code
<% Class.forName("sun.jdbc.odbc.JdbcOdbcDriver"); Connection con = DriverManager.getConnection("jdbc:odbc:tscgdb","sa","sa"); Statement stat = con.createStatement(); ResultSet rs = stat.executeQuery("select * from books"); while(rs.next()){ String name = rs.getString("bookname"); %> <option value="<%=name%>"> <%=name%></option> <% } rs.close(); %>