Java 基于android中某些条件的ArrayList过滤器
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ArrayList filter based on some condition in android
提问by user3061048
I have an ArrayListin android which contains several names. Suppose user typed R, so I want to calculate how many names are there in the ArrayListwhich starts with R. for example if arraylist contains Ashok, Bimal, Ram, Raju, sunita then it should return 2 as there are 2 names starts with R
我有一个ArrayListin android，其中包含多个名称。假设用户输入了 R，所以我想计算其中有多少个名字ArrayList以 R 开头。例如，如果 arraylist 包含 Ashok、Bimal、Ram、Raju、sunita，那么它应该返回 2，因为有 2 个名字以 R 开头
采纳答案by user543
int count = 0;
ArrayList<String> _list;//ur arraylist with names
for (String names : _list) {
if (names.startsWith("R")) {
count++;
}
}
System.out.println(count);
回答by doorstuck
Iterate through the ArrayList and increment an integer variable when your check returns true.
遍历 ArrayList 并在检查返回 true 时增加一个整数变量。
回答by Gaurav Gupta
ArrayList<String> list = new ArrayList<String>();
list.add("Ashok");
list.add("Bimal");
list.add("Ram");
list.add("Raju");
list.add("sunita");
int count = 0;
for(String s: list){
if(s.startsWith("R")){
count++;
}
}
System.out.println(count);
回答by DroidDev
In your onCreate method
在您的 onCreate 方法中
int countToShow=counterMethod(character);
TextView showCount=(TextView)findViewById(idOfYourTextView);
showCount.setText(String.valueOf(count));
Make another method in your activity as follow:-
在您的活动中制作另一种方法，如下所示：-
public int counterMethod(String character){
int count=0;
for(int i=0; i<arrayListName.size();i++)
{
String name=arrayListName.get[i];
if(name.startswith(character))
count++;
}
return count;
}
回答by suresh
Try below code....
试试下面的代码....
enter code here
List<String> list = new ArrayList<String>();
list.add("How are you");
list.add("How you doing");
list.add("Joe");
list.add("Mike");
Collection<String> filtered = Collections2.filter(list,
Predicates.containsPattern("How"));
print(filtered);
