英文:
How to increment count within foreach loop
问题
public static void main(String[] args) {int count = 0;List<String> namesList = new ArrayList<>();namesList.add("gaurav");namesList.add("deepak");namesList.add("anit");namesList.add("garvit");namesList.add("satvir");namesList.add("lino");namesList.add("gogo");namesList.forEach(names -> {if (names.startsWith("g")) {count++;}});if (count > 1) {System.out.println("executed");}}
在增加计数时遇到以下错误：
在封闭范围中定义的局部变量 "count" 必须是 final 或者是 effectively final
英文:
Below is my code:
public static void main(String[] args) {int count = 0;List<String> namesList = new ArrayList<>();namesList.add("gaurav");namesList.add("deepak");namesList.add("anit");namesList.add("garvit");namesList.add("satvir");namesList.add("lino");namesList.add("gogo");namesList.forEach(names -> {if (names.startsWith("g")) {count++;}});if (count > 1) {System.out.println("executed");}}
Getting the below error while incrementing count:
Local variable count defined in an enclosing scope must be final or effectively final
答案1
得分: 1
不要增加一个变量。
long count = namesList.stream().filter(name -> name.startsWith("g")).count();
这样更易于阅读和理解，而且代码更简洁。
英文:
Don't increment a variable.
long count = namesList.stream().filter(name -> name.startsWith("g")).count();
It's eaisier to read and understand, and it's less code.
答案2
得分: 0
使用 AtomicInteger 进行如下操作：
AtomicInteger count = new AtomicInteger();namesList.forEach(names -> {if (names.startsWith("g")) {count.getAndIncrement();}});if (count.get() > 1) {System.out.println("executed");}
或者使用 Stream API 进行如下操作：
long count = namesList.stream().filter(name -> name.startsWith("g")) // 使用条件进行过滤.count(); // 然后计数
英文:
Use AtomicInteger for this
AtomicInteger count = new AtomicInteger();namesList.forEach(names -> {if (names.startsWith("g")) {count.getAndIncrement();}});if(count.get() > 1) {System.out.println("executed");}
Or using Stream API
long count = namesList.stream().filter(name -> name.startsWith("g")) // filter with condition.count(); // then count
答案3
得分: 0
你也可以使用流来实现这个。使用 filter() 来查找以 g 开头的名称，使用 findFirst() 来找到第一个。如果没有符合条件的，将返回 null。
String result = namesList.stream().filter(name -> name.startsWith("g")).findFirst().orElse(null);
英文:
You can do that with a stream as well. Use filter() to find the names starting with g, use findFirst(), to find the first one. If there is none, this will return null.
String result = namesList.stream().filter(name -> name.startsWith("g")).findFirst().orElse(null);
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