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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?
我基本上是一個php程序員,現在只是想在Java的海岸上徘徊(或徘徊!)。 當然是救生艇,stackoverflow。
我正在努力在Java中制作多維數組,就像在php中那樣,完全有可能$array["bla"]["blabla"]["blablabla"]...
這是我想要實現的
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"));
有人請幫助我繼續前進..
編輯:使用php示例來闡明所需的數組
Java中這樣的代碼的優良作法是不要為此使用無類型的數組,而要制作實際的有類型的對象:
class Whatever {
private final String username;
private final List<Group> groups;
...
}
class Group {
...
}
像這樣定義你的對象
class SampleModel{
String userName;
List<String> groups ;
List<String> categories;
String notification;
List<String> departments;
Map<String,List<String>> sub_deptmnt;
//getter and setter
}
如果您提前知道鍵將是什么,那么最好以Prabhakaran向您顯示的方式設置帶有字段和getter / setter方法的類。
但是,如果鍵可以是動態的:與PHP關聯數組等效的是Map<String,Object>
。
Map<String,Object> arr = new HashMap<String,Object> ();
然后創建一個數組元素:
arr.put ("user", userName);
並獲取它:
String userName = (String)(arr.get ("user"));
您可以將數組元素設置為數組, ArrayList
或其他Map<String,Object>
映射,以獲取多維數組。 事實是,Java是強類型的,這意味着get
方法將返回一個Object,並且您必須將其強制轉換為期望的類型:
String[] categories = (String[])(arr.get ("categories"));
要么
ArrayList<String> categories = (ArrayList<String>)(arr.get ("categories"));
如果先前為“類別”存儲的對象的類型錯誤,則將引發異常。
如果沒有其他問題,您至少應該研究一下collections教程 ,因為您絕對需要了解它是否可以使用Java。
[注意:我尚未測試以上任何代碼。 我想我做對了。]
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