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[英]Multi-dimensional (array of arrays) in Java. Are they row or column arrays?
[英]Multi-dimensional array in Java. Is it that complex?
我基本上是一个php程序员,现在只是想在Java的海岸上徘徊(或徘徊!)。 当然是救生艇,stackoverflow。
我正在努力在Java中制作多维数组,就像在php中那样,完全有可能$array["bla"]["blabla"]["blablabla"]...
这是我想要实现的
Array
(
[user] => UserName
[groups] => Array
(
[0] => group1
[1] => group2
[2] => group3
)
[categories] => Array
(
[0] => category1
[1] => category2
[2] => category3
)
[notification] => user notification string
[departments] => Array
(
[0] => department1
[1] => department2
[2] => department3
[3] => department4
)
[sub-deptmnt] => Array
(
[department1] => Array
(
[0] => subdep1
[1] => subdep2
)
[department2] => Array
(
[0] => another-subdep1
[1] => another-subdep2
[2] => another-subdep3
)
)
)
在PHP中
$array["user"] = "UserName";
$array["groups"] = array("group1","group2","group3");
$array["categories"] =array("category1","category2","category3");
$array["notification"] = "user notification string";
$array["departments"] = array("department1","department2","department3","department4");
$array["sub-deptmnt"] = array("department1" => array("subdep1","subdep2"),"department2"=> array("another-subdep1","another-subdep2", "another-subdep3"));
有人请帮助我继续前进..
编辑:使用php示例来阐明所需的数组
Java中这样的代码的优良作法是不要为此使用无类型的数组,而要制作实际的有类型的对象:
class Whatever {
private final String username;
private final List<Group> groups;
...
}
class Group {
...
}
像这样定义你的对象
class SampleModel{
String userName;
List<String> groups ;
List<String> categories;
String notification;
List<String> departments;
Map<String,List<String>> sub_deptmnt;
//getter and setter
}
如果您提前知道键将是什么,那么最好以Prabhakaran向您显示的方式设置带有字段和getter / setter方法的类。
但是,如果键可以是动态的:与PHP关联数组等效的是Map<String,Object>
。
Map<String,Object> arr = new HashMap<String,Object> ();
然后创建一个数组元素:
arr.put ("user", userName);
并获取它:
String userName = (String)(arr.get ("user"));
您可以将数组元素设置为数组, ArrayList
或其他Map<String,Object>
映射,以获取多维数组。 事实是,Java是强类型的,这意味着get
方法将返回一个Object,并且您必须将其强制转换为期望的类型:
String[] categories = (String[])(arr.get ("categories"));
要么
ArrayList<String> categories = (ArrayList<String>)(arr.get ("categories"));
如果先前为“类别”存储的对象的类型错误,则将引发异常。
如果没有其他问题,您至少应该研究一下collections教程 ,因为您绝对需要了解它是否可以使用Java。
[注意:我尚未测试以上任何代码。 我想我做对了。]
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