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，如果我想得到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"];
        }
        ?>

......有一个错误：

致命错误：未捕获PDOException：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法有错误;检查手册，对应右边的MySQL服务器版本语法使用附近＆＃39; dog.id AS d FROM dog LEFT JOIN cat ON dog.name = cat.name＆＃39;在/myproject.php：31的第3行堆栈跟踪：＃0 /myproject.php(31）：PDOStatement-＆gt;执行（数组）＃1 {main}抛出第31行的/myproject.php

Answer 1

根据您的数据和错误消息，horse.id AS h之后您的查询中错过了一个逗号

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

Answer 2

你错过了一个逗号：

    SELECT *,
    horse.id AS h  -- HERE
    dog.id AS d

应该是：

    SELECT *,
    horse.id AS h,
    dog.id AS d

如何获得左边加入mySQL表的id？

2 个答案: