php - 如何获取左连接 mySQL 表的 id？

Question

已关闭。这个问题是 not reproducible or was caused by typos 。目前不接受答案。

---

这个问题是由拼写错误或无法再重现的问题引起的。虽然类似的问题可能是 on-topic在这里，这个问题的解决方式不太可能帮助 future 的读者。

已关闭 6 年前。

Improve this question

像这样，我的 mySQL 请求运行良好:

 <?php
       $id = 12;

        $pdo = $db->prepare("
        SELECT *,
        horse.id AS h
        FROM dog 
        LEFT JOIN cat ON dog.name=cat.name
        LEFT JOIN horse ON dog.name=horse.name
        WHERE animal = ?");
        $pdo->execute(array($id));
        $animals = $pdo->fetchAll();

        foreach ($animals as $row) {
            echo $row["h"];
        }
        ?>

但是如果我想获取dog.id...

 <?php

        $id = 12;

        $pdo = $db->prepare("
        SELECT *,
        horse.id AS h
        dog.id AS d
        FROM dog 
        LEFT JOIN cat ON dog.name=cat.name
        LEFT JOIN horse ON dog.name=horse.name
        WHERE animal = ?");
        $pdo->execute(array($id));
        $animals = $pdo->fetchAll();

        foreach ($animals as $row) {
            echo $row["h"];
            echo $row["d"];
        }
        ?>

...有错误:

Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'dog.id AS d FROM dog LEFT JOIN cat ON dog.name=cat.name' at line 3 in /myproject.php:31 Stack trace: #0 /myproject.php(31): PDOStatement->execute(Array) #1 {main} thrown in /myproject.php on line 31

Answer 1

根据您的数据和错误消息，您在查询中在 horse.id AS h 之后漏掉了一个逗号，它应该是

SELECT *,
 horse.id AS h,
 dog.id AS d
FROM dog 
LEFT JOIN cat ON dog.name=cat.name
LEFT JOIN horse ON dog.name=horse.name