我有一个算法，该算法具有一个名为<head runat="server">
<title></title>
<script src="Scripts/jquery-1.10.2.min.js"></script>
<script type="text/javascript">
!function ($, window) {
$(function () {
init();
});
var getAddresses = function (postcode, callback) {
//var $xhr = $.getJSON('/lookupPostcode/' + postcode);
//$xhr.done(function (data) {
// callback(data);
//});
//$xhr.error(function () {
// callback([]);
//})
//using the following array to populate the select.
var datalist = [{ "ID": "1", "ParentID": "0", "Name": "Parent1" }, { "ID": "2", "ParentID": "0", "Name": "Parent2" },
{ "ID": "3", "ParentID": "1", "Name": "Child 1.1" }, { "ID": "4", "ParentID": "1", "Name": "Child 1.2" },
{ "ID": "5", "ParentID": "3", "Name": "SubChild 1.1.1" }, { "ID": "6", "ParentID": "2", "Name": "Child 2.1" }];
callback(datalist);
};
var init = function () {
$("input.address").each(function () {
var $input = $(this);
var $icon = $input.next('.address-button');
//you are using the .row class to find the select control, but from your form, I can't find this class.
var $select = $input.parents('.row').eq(0).find("select");
//debugger;
$select.css("display", "none");
var onKeyUp = function (e) {
if (e.keyCode === 13) {
e.preventDefault();
e.stopPropagation();
$icon.trigger("click");
}
};
var onKeyDown = function (e) {
if (e.keyCode === 13) {
e.preventDefault();
e.stopPropagation();
}
};
$input.on("keyup", onKeyUp);
$input.on("keydown", onKeyDown);
$icon.on("keyup", onKeyUp);
$icon.on("keydown", onKeyDown);
$select.on("keyup", onKeyUp);
$icon.on("click", function () {
$select.empty();
$select.append("<option value=''>Enter above</option>");
getAddresses($input.val(), function (addresses) {
//populate select options with addresses
$.each(addresses, function (index, item) {
//debugger;
$select.append("<option value='" + item.ID + "'>" + item.Name + "</option>");
});
$select.css("display", "block");
});
});
$select.on('change', function (event) {
var ua = window.navigator.userAgent;
var is_ie = /MSIE|Trident/.test(ua);
//get the selected text and populate the input text.
$input.val($(this).find("option:selected").text());
//hide the select control
if (!is_ie) {
$select.css("display", "none");
}
$select.css("display", "none");
});
});
};
}(jQuery, window);
</script>
</head>
<body>
<form action="http://localhost:8000/processForm" method="post">
<label for="input_1" class="form-control-label col-xs-12">
Postcode
</label>
<div class="col-xs-12 row">
<div class="input-group">
<input type="text" name="questions[1]" autocomplete="off" id="input_1" class="form-control address" value="" />
<a class="btn btn-xs input-group-addon address-button" id="input_address_addon" role="button" tabindex="0">
<img src="http://i.imgur.com/H9FIags.jpg" style="height:10px;width:10px" />
Search
</a>
</div>
<div class="col-xs-12 col-sm-8 col-sm-offset-4 col-md-7">
<select class="form-control selectpicker addSelect" id="input_address_select" >
<option value="">Enter above</option>
</select>
</div>
</div>
<button type="submit" class="btn submit btn-navigation">
Continue
</button>
</form>
</body>
的通用类以及一些从Employee继承的特定类，如下所示：
Employee我正在寻找一种避免单独调用每个对象并从继承自especific类的abstract class Employee {
private String name;
private String document;
public String toString(){
return "\nName and Document";
}
}
class Driver extends Employee {
private String licencePlate;
public String toString(){
return "Plate: XXXXX";
}
}
class TelOperator extends Employee {
private String phone;
public String toString(){
return "Phone: XXX - XXXXXX";
}
}
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
TelOperator telOp = new TelOperator();
System.out.println(telOp);
Driver driver = new Driver();
System.out.println(driver);
}
}
执行主类的toString方法的方法。
我的输出将是这样的：
//驱动程序
//电话接线员
您可以检查源代码here。
答案 0 :(得分：0)
只需在子类的super.toString()方法中添加对toString()的调用：-
public String toString() {
return super.toString() + "\nPhone: XXX - XXXXXX";
}
答案 1 :(得分：0)
如果要一次打印两个子类，则必须创建一个子类的对象并在主对象的toString中调用它们的toString。希望我能正确理解您的问题。
答案 2 :(得分：0)
您可以为员工创建一个Factory类，然后工厂决定应将哪个子类返回给调用程序。然后，它将创建该子类的实例，然后将其返回到调用程序。
希望我能正确理解您的问题。