Flash 字典()as3
我使用字典变量来匹配拖放练习Flash 字典()as3,flash,actionscript-3,Flash,Actionscript 3,我使用字典变量来匹配拖放练习 var dict = new Dictionary (); dict[box_a]=s1; dict[box_b]=s2; dict[box_c]=s3; dict[box_d]=s4; 问题1:最后,我想检查带有s1的复选框a==是否,依此类推。。。。我该怎么做 for each( var item in dict) { item.removeEventListener(MouseEvent.MOUSE_DOWN, mouseDownHandler
var dict = new Dictionary ();
dict[box_a]=s1;
dict[box_b]=s2;
dict[box_c]=s3;
dict[box_d]=s4;
问题1:最后,我想检查带有s1的复选框a==是否,依此类推。。。。我该怎么做
for each( var item in dict)
{ item.removeEventListener(MouseEvent.MOUSE_DOWN, mouseDownHandler);
item.removeEventListener(MouseEvent.MOUSE_UP, mouseUpHandler);
if(item==????)
// what do i have to put instead of ???
hits=hits+1;
问题2:我希望box_d也接受s3,我会怎么做
如果我这样做
dict[box_d]=s4;和dict[box_d]=s3;它不起作用,因为它与最后一个相等项相匹配(
谢谢!!似乎只有基元类型才能检查相等性,用
==
替换=
运算符不会改变任何内容:
var dict:Dictionary = new Dictionary();
dict['ololo1'] = { one : 1, two : '2' };
dict[ { one : 1, two : '2' } ] = function(o:*):String { return o as String; };
dict[function(o:*):String { return o as String; }] = 'ololo';
dict['ololo2'] = 'ololo2';
dict[{ one : 1, two : '2' }] = { one : 1, two : '2' };
for (var key:* in dict) {
trace(key, dict[key], key == dict[key]);
}
trace('equal functions :', (function(o:*):String { return o as String; } ===
function(o:*):String { return o as String; } ));
trace('equal objects :', ({ one : 1, two : '2' } ===
{ one : 1, two : '2' } ));
trace('equal functions :', (function(o:*):String { return o as String; } ==
function(o:*):String { return o as String; } ));
trace('equal objects :', ({ one : 1, two : '2' } ==
{ one : 1, two : '2' } ));
var obj:Object = { one : 1, two : '2' };
var obj1:Object = { one : 1, two : '2' };
var fun:Function = function(o:*):String { return o as String; };
var fun1:Function = function(o:*):String { return o as String; };
trace(obj == { one : 1, two : '2' },
obj == obj1,
fun == function(o:*):String { return o as String; },
fun == fun1);
输出
[object Object] [object Object] false
function Function() {} ololo false
[object Object] function Function() {} false
ololo2 ololo2 true
ololo1 [object Object] false
equal functions : false
equal objects : false
equal functions : false
equal objects : false
false false false false
但是,
obj==obj
返回true
;)似乎只能对基本类型检查相等性,用==
替换=
运算符不会改变任何内容:
var dict:Dictionary = new Dictionary();
dict['ololo1'] = { one : 1, two : '2' };
dict[ { one : 1, two : '2' } ] = function(o:*):String { return o as String; };
dict[function(o:*):String { return o as String; }] = 'ololo';
dict['ololo2'] = 'ololo2';
dict[{ one : 1, two : '2' }] = { one : 1, two : '2' };
for (var key:* in dict) {
trace(key, dict[key], key == dict[key]);
}
trace('equal functions :', (function(o:*):String { return o as String; } ===
function(o:*):String { return o as String; } ));
trace('equal objects :', ({ one : 1, two : '2' } ===
{ one : 1, two : '2' } ));
trace('equal functions :', (function(o:*):String { return o as String; } ==
function(o:*):String { return o as String; } ));
trace('equal objects :', ({ one : 1, two : '2' } ==
{ one : 1, two : '2' } ));
var obj:Object = { one : 1, two : '2' };
var obj1:Object = { one : 1, two : '2' };
var fun:Function = function(o:*):String { return o as String; };
var fun1:Function = function(o:*):String { return o as String; };
trace(obj == { one : 1, two : '2' },
obj == obj1,
fun == function(o:*):String { return o as String; },
fun == fun1);
输出
[object Object] [object Object] false
function Function() {} ololo false
[object Object] function Function() {} false
ololo2 ololo2 true
ololo1 [object Object] false
equal functions : false
equal objects : false
equal functions : false
equal objects : false
false false false false
但是,obj==obj
返回true
;) 介绍
字典是键/值的集合,类似于表:
+------+--------+
| key1 | value1 |
+------+--------+
| key2 | value2 |
+------+--------+
| key3 | value3 |
+------+--------+
| key4 | value4 |
+------+--------+
要访问值,必须使用键,即:dict[key2]
返回value2
答复1:
您不能直接访问密钥,但可以使用for()
语句访问密钥,因此:
for (var key:* in dict) {
trace('dict[' + key + '] = ' + dict[key]);
}
for each (var value:* in dict) {
trace(value);
}
将输出以下内容:
dict[key1] = value1
dict[key2] = value2
dict[key3] = value3
dict[key4] = value4
value1
value2
value3
value4
对于每个()语句,您都使用了一个。对于该特定语句,您将得到另一个结果,因为您正在迭代值而不是键,因此:
for (var key:* in dict) {
trace('dict[' + key + '] = ' + dict[key]);
}
for each (var value:* in dict) {
trace(value);
}
将输出以下内容:
dict[key1] = value1
dict[key2] = value2
dict[key3] = value3
dict[key4] = value4
value1
value2
value3
value4
因此,如果您想使用密钥检查某些内容,必须使用for(key in dict)
表单
答复2:
我建议您使用如下数组:
dict[box_a] = [s1];
dict[box_b] = [s2];
dict[box_c] = [s3];
dict[box_d] = [s3, s4];
然后以数组形式访问这些值,并检查所有值:
for (var key:* in dict) {
var values:Array = dict[key];
for each (var value:* in values) {
// do what you want with s1, s2, s3, etc
}
}
简介
字典是键/值的集合,类似于表:
+------+--------+
| key1 | value1 |
+------+--------+
| key2 | value2 |
+------+--------+
| key3 | value3 |
+------+--------+
| key4 | value4 |
+------+--------+
要访问值,必须使用键,即:dict[key2]
返回value2
答复1:
您不能直接访问密钥,但可以使用for()
语句访问密钥,因此:
for (var key:* in dict) {
trace('dict[' + key + '] = ' + dict[key]);
}
for each (var value:* in dict) {
trace(value);
}
将输出以下内容:
dict[key1] = value1
dict[key2] = value2
dict[key3] = value3
dict[key4] = value4
value1
value2
value3
value4
对于每个()
语句,您都使用了一个。对于该特定语句,您将得到另一个结果,因为您正在迭代值而不是键,因此:
for (var key:* in dict) {
trace('dict[' + key + '] = ' + dict[key]);
}
for each (var value:* in dict) {
trace(value);
}
将输出以下内容:
dict[key1] = value1
dict[key2] = value2
dict[key3] = value3
dict[key4] = value4
value1
value2
value3
value4
因此,如果您想使用密钥检查某些内容,必须使用for(key in dict)
表单
答复2:
我建议您使用如下数组:
dict[box_a] = [s1];
dict[box_b] = [s2];
dict[box_c] = [s3];
dict[box_d] = [s3, s4];
然后以数组形式访问这些值,并检查所有值:
for (var key:* in dict) {
var values:Array = dict[key];
for each (var value:* in values) {
// do what you want with s1, s2, s3, etc
}
}
什么是s*变量(s1、s2等)?您是否没有在代码中保留对它们的引用(您正在创建它们,并且没有将引用存储在dict
变量之外的任何位置)?你能为你正在做的事情提供更多的上下文(更多的代码)吗?s*变量(s1、s2等)是什么?您是否没有在代码中保留对它们的引用(您正在创建它们,并且没有将引用存储在dict
变量之外的任何位置)?你能为你正在做的事情提供更多的上下文(更多的代码)吗?非常感谢你的解释!!如果我需要查看框a中的对象是否为s1,我会怎么做?对于(var key:*in dict){If(key==dict[key]){hits=hits+1;}}}P.s这些键和dict[key]都是movieclipsy,你应该这样做:key==dict[key],如果你想知道键的值是否就是键本身。谢谢你的精彩解释!!如果我需要查看框a中的对象是否为s1,我会怎么做?对于(var key:*in dict){If(key==dict[key]){hits=hits+1;}}}P.s这些键和dict[key]都是movieclipsys,如果你想知道键的值是否是键本身,你应该这样做:key==dict[key]。