java 仅从 JavaFX ObservableList 中获取一些值
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Take only some value from JavaFX ObservableList
提问by Alfat Saputra Harun
I need to build a
我需要建立一个
LineChart<Number,Number>
I had the data stored in
我将数据存储在
ObservableList<MyData>
MyDatahad 4 variable, all of it are int. Let just say the variable in MyDataare: No1, No2, No3, No4.
MyData有 4 个变量，它们都是int. 假设变量MyData是： No1, No2, No3, No4。
Next, I need to build
接下来，我需要构建
LineChart<Number,Number>
and I need only No1and No2variable, but I don't know how to take that value from ObservableList, Now I just use XYChart.Datato add new data to XYChart.Seriesof my LineChart<Number,Number>like this:
而我只需要No1和No2变化，但我不知道如何把从该值ObservableList，现在我只是用XYChart.Data新的数据添加到XYChart.Series我的LineChart<Number,Number>是这样的：
private static XYChart.Series dataLineChart = new XYChart.Series();
public static void updateDataChart(){
dataLineChart.getData().addAll(
new XYChart.Data(3,15),
new XYChart.Data(7,20)
);
}
If only I can take the value from ObservableListI can just simply use :
如果我能从中获取价值，ObservableList我就可以简单地使用：
private static XYChart.Series dataLineChart = new XYChart.Series();
public static void updateDataChart(){
dataLineChart.getData().addAll(
myObservableList
);
}
Can someone help me with this problem ?
有人可以帮我解决这个问题吗？
回答by Sergey Grinev
You need to create a new list anyway. It's not a problem, it's fast and only a bit of new memory will be used (only for handlers, not for objects):
无论如何，您都需要创建一个新列表。这不是问题，它很快，并且只会使用一点新内存（仅用于处理程序，而不用于对象）：
If you know positions of required items you can use List#subList()method.
如果您知道所需物品的位置，则可以使用List#subList()方法。
private static XYChart.Series dataLineChart = new XYChart.Series();
public static void updateDataChart(){
dataLineChart.getData().addAll(
myObservableList.subList(0,2);
);
}
If items are not consequent, you may to create new list in one line:
如果项目不是结果，您可以在一行中创建新列表：
ObservableList<String> sublist = FXCollections.observableArrayList(
myObservableList.get(3), myObservableList.get(5) );
Also you may save on new list if you use your condition to filter old list into new directly in update method:
如果您使用条件直接在更新方法中将旧列表过滤为新列表，您也可以保存在新列表中：
public static void updateDataChart(){
for (MyData data : myObservableList) {
if ( data.isLucky() ) // or whatever is your condition
dataLineChart.getData().add(data);
}
}
