Question

我的数据库中有两个表。其中一个是名为players的表，另一个是bans：

players table: ID, Score
bans table: user_id, reason

我需要做的是：选择原因=作弊和得分<250 。

当我尝试FULL JOIN时我尝试做一些JOINS，我发现MySQL中不存在该东西，所以任何帮助都会很好，谢谢！！

我也这样尝试过，但我总是得到零行

SELECT bans.user_id, players.ID
FROM bans
INNER JOIN players ON bans.user_id=players.ID;

禁令表

user_id      reason
133032       swearing
133040       name not allowed

玩家桌

id            score
15            13378
21            215216
133032        15
133040        157

Answer 1

您的查询是一个好的开始。对于您的样本数据，它将为您提供结果。为了满足您的要求，您只需要几个其他过滤器：

public static void main(String[] args) {

    int maxContacts = 20;
    Contact[] contacts = new Contact[maxContacts];    // initialize array of size 20
    Scanner scanner = new Scanner(System.in);
    String firstName;
    String lastName;
    String phone;
    String email;

   //collect 20 contacts into array
    for (int i = 0; i < maxContacts; i++) {

         System.out.println("insert first name : ");
         firstName = scanner.nextLine();
         System.out.println("insert last Name : ");
         lastName = scanner.nextLine();
         System.out.println("insert phone : ");
         phone = scanner.nextLine();
         System.out.println("insert email : ");
         email = scanner.nextLine();

         contacts[i] = new Contact(firstName, lastName, phone, email);
    }

   // Just print contacts array is full
    if (contacts.length == maxContacts) {
        System.out.println("maxiumum number of adding contact has reached");
        }

      //Now just sort the array based on name
      Arrays.sort(contacts);

     //Finally print contacts in order
     for (Contact contact : contacts) {
        System.out.println(contact.toString());

    }

 }

我无法绕过MySQL中的INNER JOIN

1 个答案: