Question

我有2张桌子

rankID |名称


1 |新

2 |学习者

3 |经历

4 |亲

另一个包含所有用户信息，密码和内容

id |用户名| rankID


1 |你好| 3

2 |嘿| 3

我到目前为止所以我可以显示他们的排名，但我想显示排名。我怎样才能做到这一点？我已经尝试了很多东西但是我不太擅长sql和它的php部分。

这是我用来显示排名编号的代码

//Get rankID
$query = "SELECT rankID FROM users WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rankID'];

并显示等级编号：

Rank: <?php echo $rank; ?>

Answer 1

简单JOIN查询： -

"SELECT rank.name as rank_name,users.rankID as rankID from users LEFT JOIN rank ON rank.rankID = users.rankID WHERE id = '$userId'"

然后： -

$query = "SELECT rank.name as rank_name,users.rankID as rankID from users LEFT JOIN rank ON rank.rankID = users.rankID WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);

执行： -

$rank = $row['rankID'];
$rank_name = $row['rank_name'];

Rank: <?php echo $rank; ?>
RankName: <?php echo $rank_name; ?>

或者

$rank_data = $row;
Rank: <?php echo $rank_data['rankID']; ?>
RankName: <?php echo $rank_data['rank_name']; ?>

不是： - 其他程序员在评论和回答中列出了许多其他可能的方式。

Answer 2

//获取数据

$query = "SELECT rankID, name FROM users WHERE id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rankID'];
$rank = $row['name'];

并显示数据：

Rank: <?php echo $rank; ?>
Name: <?php echo $name; ?>

Answer 3

希望这对你有用。

$query = "SELECT rankID FROM users WHERE id = '".$userId."'";
$result = $conn->query($query);
$count = $result->num_rows;
if($count==0)
{
   return false;
}
else
{
     $rows=[];
     while($row = $result->fetch_assoc())
     {
        $rows[] = $row;
     }
     return $rows;
}

Answer 4

请使用以下代码

$query = "SELECT name FROM users as u JOIN rank as r ON r.rankID = u.rankID WHERE u.id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$name = $row['name'];


Name: <?php echo $name; ?>

Answer 5

试试这个：

//Get rankID
$query = "SELECT rankID, rank.name AS rank_name FROM rank, users WHERE id = '$userId' and users.rankid = rank.rankid";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rank_name'];
echo $rank;

Answer 6

当您想从两个不同的表中获取数据时。您需要加入查询。以下是您的查询，它将明确解决您的问题：

$q="select a.name,b.rankID from rankname as a INNER JOIN user as b
ON a.rankID = b.rankID";

有关如何加入两个表的更多信息，请参阅：http://www.tutorialspoint.com/sql/sql-using-joins.htm

希望这会对你有所帮助。

Answer 7

请试试这个

//Get rankID
$query = "SELECT r.name as rank_name FROM rank as r inner join users as u on r.rankID = u.rankID WHERE u.id = '$userId'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_assoc($result);
$rank = $row['rank_name'];

echo 'Rank: '. $rank;

SQL - 显示名称而不是ID

7 个答案: