使用内连接 SQL 语句时未定义的索引

Tob*_*ith 2 mysql sql pdo join

嗨,我正在尝试编写 SQL 语句来填充表,但我不断收到错误消息:

未定义索引:st.Name,和未定义索引:s.Name。

我不明白为什么我会得到它,因为我在 SQL 语句中选择了它们。我不太擅长 SQL,因此将不胜感激。

<?php
ini_set("display_errors", 1);
try{
    $stmt = $conn->prepare(
        "SELECT st.Name, s.Name
        From Sports AS s INNER JOIN Choices AS c
        ON s.Sport_ID = c.Sport_ID INNER JOIN Student_Choices AS sc
        ON sc.T1_Choice = c.Choice_ID INNER JOIN Students AS st
        ON st.Username = sc.Username
    ");
    $stmt->execute();
    while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
        echo '<tr>
                <td>'.$row['st.Name'].'</td>
                <td>'.$row['s.Name'].'</td>
         </tr>
        ';
    }
}
catch(PDOException $e)
{
    echo "error".$e->getMessage();
}
?>
Run Code Online (Sandbox Code Playgroud)

Mad*_*iya 5

有多个问题。首先,您假设您将能够使用<table/alias-name>.<column/alias-name>. 相反,它们只能使用列名或定义的别名来访问。

现在,在这种情况下,您有两个名称相同的列。因此,您应该为它们定义不同的别名,以避免产生歧义。我已经定义了它们student_namesport_name.

定义别名后,您现在可以仅使用别名访问这些列值。

        $stmt = $conn->prepare(
          "SELECT st.Name AS student_name, 
                  s.Name AS sport_name 
          From Sports AS s INNER JOIN Choices AS c
          ON s.Sport_ID = c.Sport_ID INNER JOIN Student_Choices AS sc
          ON sc.T1_Choice = c.Choice_ID INNER JOIN Students AS st
          ON st.Username = sc.Username
        ");
        $stmt->execute();
        while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
          echo '<tr>
              <td>'.$row['student_name'].'</td>
              <td>'.$row['sport_name'].'</td>
            </tr>
          ';
        }
Run Code Online (Sandbox Code Playgroud)