Question

我有两个名为'user'的表，另一个名为'user_info'。

用户;

userid       |      username
------------------------------
1            |      mary
2            |      john
3            |      liam

USER_INFO;

userid       |      desc
----------------------------------------------------
1            |      hello, my name is mary i am 26
2            |      message me if you need any help
3            |      please leave me alone

我想根据userID从'user_info'表中显示来自'user'表的信息的相应描述。所以它看起来像：

Name: John
UserID: 2
Description: message me if you need any help

这是我的代码;

//Query the database
$resultSet = $mysqli->query("SELECT * FROM user,user_info");

//Count the returned rows
if ($resultSet->num_rows != 0) {


//Turn the results into an ArrayAccess
while($rows = $resultSet->fetch_assoc())
{
    $uid = $rows['UserID'];
    $username = $rows['username'];
    $desc = $rows['desc'];

echo "<p>Name: $username </p>
      <p>User ID: $uid </p>
      <p>Description: $desc </p>
      <hr>";
}

目前它会尽可能多地返回不同的组合;像这样：

Name:Mary
UserID:1
Description:hello, my name is mary i am 26
--------------------------------------------
Name:John
UserID:1
Description:message me if you need any help
--------------------------------------------
Name:Liam
UserID:1
Description:please leave me alone
--------------------------------------------
Name:Mary
UserID:2
Description:hello, my name is mary i am 26
--------------------------------------------
Name:John
UserID:2
Description:message me if you need any help
--------------------------------------------
ETC...

如何才能使它只返回基于userID的正确相应数据（仅一次），以便它显示所有用户的正确用户名，ID和描述？

我不知道怎么解释这个，如果有人希望我进一步详细说明我再试一次没问题，

Answer 1

更新您的请求

$resultSet = $mysqli->query("SELECT * FROM user 
    JOIN user_info ON user_info.userid = user.userid");

Answer 2

尝试修改您的查询，如下所示：

SELECT user.*,user_info.desc FROM user INNER JOIN user_info ON user.userid=user_info.userid;

Answer 3

像这样建立内部联接查询：

Select userId, userName, desc from user inner join user_info on
user.userid = user_info.userid

这应该可以解决问题。

Answer 4

您可以使用join方法根据它们的关系从不同的表中获取数据，因为您的问题查询将是这样的，

select user.username as Name, user.userid as UserID, user_info.desc as Description from user inner join user_info on user.userid=user_info.userid;

PHP MySQL从2个不同的表中选择*并显示合并的数据

4 个答案: