Question

在我的数据库中，我有3个表。

用户（表名）如此多的行是用户名之一，如果

--------------------
| id | user        |
--------------------
| 7  | user        |
| 8  | user_name_2 |
| 11 | user_name_5 |
--------------------

和另一个表格调用数据

----------------------------
| id | user-id | number    |
----------------------------
| 1  | 7       | 789654125 |
| 2  | 8       | 465654545 |
| 3  | 11      | 884554511 |
----------------------------

在表td中，user_id是表用户ID

现在我想在php中显示姓名和号码

$conn = mysqli_connect("localhost","root","QAZWS12","user");
$sql =  "SELECT * FROM user"  
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
    $id = $row["id"];
}

相同并获得$number = $row["number"]

现在我要在数据表用户ID中自动获取表用户号，我该怎么做？

最终输出

用户789654125 像这样的user_name_2 465654545

Answer 1

You have to use join here. For more details on different types JOINS refer either official MySQL doc or search online

replace this query

$sql = "SELECT u.id,u.user AS user_name,d.number AS user_number 
FROM user u LEFT JOIN  description d ON  u.id = d.user-id";

The above query will fetch user id and username and user number

replace this too

$id = $row["id"].' '.$row['user_name'].' '.$row['user_number'].'<br/>';

This outputs as 7 user 789654125

Answer 2

Try below SQL query with LEFT JOIN to get the username with their mobile number in the result set:

$sql =  "SELECT u.user, d.number 
         FROM user as u 
         LEFT JOIN data as d ON d.user-id = u.id
        "

Answer 3

用户SQL左联接

$sql = "SELECT * FROM user LEFT JOIN call_data ON user.id = call_data.user-id";

访问数据时

$number = $row['number'];

了解SQL JOINS TUTORIAL

从一个标签中获取ID并找到该ID的名称

3 个答案: