Question

我有一个mysql请求，工作正常：

$pdo = $db->query('
SELECT *, 
GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
monkey.id AS id, 
duck.id AS horse_id, 
bird.id AS elephant_id
FROM monkey
LEFT JOIN duck ON monkey.horse=duck.id 
LEFT JOIN bird ON monkey.elephant=bird.id
LEFT JOIN cat ON monkey.id=cat.mouse_id 
GROUP BY monkey.id ASC;');

但现在我想只选择monkey =＆＃34; id＆＃34;。

的值

所以我尝试制作这个解决方案：

 $pdo = $db->query('
    SELECT *, 
    GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
    monkey.id AS id, 
    duck.id AS horse_id, 
    bird.id AS elephant_id
    FROM monkey WHERE id = "'.$id.'"
    LEFT JOIN duck ON monkey.horse=duck.id 
    LEFT JOIN bird ON monkey.elephant=bird.id
    LEFT JOIN cat ON monkey.id=cat.mouse_id 
    GROUP BY monkey.id ASC;');

但是我收到了这个错误：

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服务器版本要使用的语法附近＆＃39; LEFT JOIN duck on monkey.chorse = duck.id LEFT JOIN 鸟在猴子.bir＆＃39;在..mypage.php的第7行：28堆栈跟踪：＃0 ..mypage.php（28）：PDO-＆gt;查询（＆＃39; \ nSELECT *，\ nGRO ...＆＃39;）＃1 {main}抛出在第28行的... mypage.php

Answer 1

首先，你必须制作where，然后制作$pdo = $db->query(' SELECT *, GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, monkey.id AS id, duck.id AS horse_id, bird.id AS elephant_id FROM monkey LEFT JOIN duck ON monkey.horse=duck.id LEFT JOIN bird ON monkey.elephant=bird.id LEFT JOIN cat ON monkey.id=cat.mouse_id WHERE monkey.id = "'.$id.'" GROUP BY monkey.id ASC;');，试试这个：

helpDialog.open()

Answer 2

所以，你有几个问题。首先，你没有清理你的id（查看Bobby Tables）。其次，您将WHERE子句放在错误的位置，第三，您不会告诉SQL使用哪个ID列。

将此作为一般模式：

SELECT
  <table1>.<column1>,
  <table2>.<column2>,
  (etc)
FROM
  <table1>
  JOIN
     <table>
     ON (<table1>.<id-column> = <table2>.<id-column)
WHERE
   <table1>.<id-column> = ${id-value}

在你的情况下，

$query = 'SELECT *, 
    GROUP_CONCAT(cat.name SEPARATOR " <br> ") AS dog, 
    monkey.id AS id, 
    duck.id AS horse_id, 
    bird.id AS elephant_id
FROM monkey 
    LEFT JOIN duck ON monkey.horse=duck.id 
    LEFT JOIN bird ON monkey.elephant=bird.id
    LEFT JOIN cat ON monkey.id=cat.mouse_id 
WHERE monkey.id = "'.$db->escape($id).'"
GROUP BY monkey.id ASC';

如何将WHERE与JOIN LEFT一起使用？

2 个答案: