在2D数组中搜索关键字

时间:2013-09-20 10:52:27

标签: java search

我正在尝试制作一个允许用户将犯罪输入“犯罪数据库”的小方法,它只会显示犯下此罪行的罪犯。

不幸的是,我必须使用2D数组。我写了一个方法,但它没有给我我正在寻找的输出。

以下是方法:

    //method to search data
public static void searchData() throws IOException{

    int flag = 0;
    boolean found = false;

//input search key
    System.out.print("Which crime would you like to select? (arson, theft, assault) ");
    String searchKey = br.readLine();

    System.out.println("You searched for criminals with the offence of \"" + searchKey + "\".");    
    System.out.println("Name - Crime - Year of Conviction");

    for(int i = 0; i < criminals.length; i++){
        if(searchKey.compareTo(criminals[i][1]) == 0){
            flag = i;
            found = true;
        }
    }

    if(found == false){
        System.out.println("Error! Crime not found.");
    }else{
        System.out.println("Criminals found.");
        for(int i = 0; i < criminals.length; i++){
            System.out.println(criminals[flag][0] + " - " + criminals[flag][1] + " - " + criminals[flag][2]);
        }
    }
} 

我的意见是:

George - theft - 1999
Eddie - assault - 2003
Al - theft - 1999

以下是测试后的输出:

Which crime would you like to select? (arson, theft, assault) theft
You searched for criminals with the offence of "theft".
Criminals found.
Al - theft - 1999
Al - theft - 1999
Al - theft - 1999

你能帮我弄清楚这有什么问题吗?提前致谢。 :)

2 个答案:

答案 0 :(得分:1)

你一直在打印最后被发现的罪犯(旗帜)。

for(int i = 0; i < criminals.length; i++){
    if(searchKey.compareTo(criminals[i][1]) == 0){
        if(!found) {
            found = true;
            System.out.println("Criminals found.");
        }
        System.out.println(criminals[i][0] + " - " + criminals[i][1] + " - " + criminals[i][2]);
    }
}

if(found == false){
    System.out.println("Error! Crime not found.");
}

答案 1 :(得分:1)

只是增加代码重用和可维护性(如果需要)的一些小增加:

  • 你已经明白在许多现实生活中,使用数组并不是最好的(=最“可读”)解决方案,因为符号(自然语言)通常比普通数字更好理解 - 编辑:在此使用常量的情况可能已经比文字数字好
  • 尽可能使用增强型for循环代替显式元素寻址:What is the syntax of enhanced for loop in Java?
  • 而不是与false进行比较,您可能希望使用!运算符来使代码更具可读性
  • 另外,.equals.compareTo .. == 0更清晰(至少在这种情况下)