在java中ArrayList中搜索 [英] search in java ArrayList

问题描述

我试图找出它的ID号来搜索在的ArrayList 客户的最佳途径。下面的code不工作;编译器告诉我，我缺少一个收益语句。

 客户findCustomerByid（INT ID）{
    布尔存在= FALSE;    如果（this.customers.isEmpty（））{
        返回null;
    }    的for（int i = 0; I＆LT; this.customers.size（）;我++）{
        如果（this.customers.get（ⅰ）.getId（）== id）的{
            存在= TRUE;
            打破;
        }        如果（存在）{
            返回this.customers.get（ID）;
        }其他{
            返回this.customers.get（ID）;
        }
    }}// Customer类是类似的东西
公共类客户{
    //属性
    INT ID;
    INT联系电话;
    串FNAME;
    字符串LNAME;
    串resgistrationDate;
}
 

解决方案

编译器是抱怨，因为您目前内部的for循环如果（存在）'块。它需要外界的吧。

 的for（int i = 0; I＆LT; this.customers.size（）;我++）{
        如果（this.customers.get（ⅰ）.getId（）== id）的{
            存在= TRUE;
            打破;
        }
}如果（存在）{
    返回this.customers.get（ID）;
}其他{
    返回this.customers.get（ID）;
}
 

话虽这么说，有更好的方法来执行此搜索。就个人而言，如果我用一个ArrayList，我的解决方案将看起来像乔恩斯基特已发布一。

I'm trying to figure out the best way to search a customer in an ArrayList by its Id number. The code below is not working; the compiler tells me that I am missing a return statement.

Customer findCustomerByid(int id){
    boolean exist=false;

    if(this.customers.isEmpty()) {
        return null;
    }

    for(int i=0;i<this.customers.size();i++) {
        if(this.customers.get(i).getId() == id) {
            exist=true;
            break;
        }

        if(exist) {
            return this.customers.get(id);
        } else {
            return this.customers.get(id);
        }
    }

}

//the customer class is something like that
public class Customer {
    //attributes
    int id;
    int tel;
    String fname;
    String lname;
    String resgistrationDate;
}

解决方案

The compiler is complaining because you currently have the 'if(exist)' block inside of your for loop. It needs to be outside of it.

for(int i=0;i<this.customers.size();i++){
        if(this.customers.get(i).getId() == id){
            exist=true;
            break;
        }
}

if(exist) {
    return this.customers.get(id);
} else {
    return this.customers.get(id);
}

That being said, there are better ways to perform this search. Personally, if I were using an ArrayList, my solution would look like the one that Jon Skeet has posted.

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