笨MySQL查询不工作

我试图让这个查询CI中工作，但它吐出一个错误：笨MySQL查询不工作

A Database Error Occurred Error Number: 1096

No tables used

SELECT *

如果我把查询直接到MySQL它工作正常（我更换$带有值的变量）。下面是该查询作为输入到笨：

$this->db->query(" 
    SELECT * 
    FROM writing_quests 
    LEFT OUTER JOIN members_quests_completed 
    ON members_quests_completed.quest_id = writing_quests.id 
    WHERE writing_quests.level_required <= $userlevel 
    AND ( 
     members_quests_completed.user_id = $user_id 
     OR 
     members_quests_completed.user_id IS NULL)" 
); 
$query = $this->db->get();

上午我做的事情是错误的，我失踪了吗？我已经包含了整个函数调用来防止问题在其他地方？我已经多次完成了这个确切的事情，没有问题。

function get_all_quests_for_user() { 
    $user_id = $this->session->userdata('user_id'); 
    $userlevel = $this->session->userdata('user_level'); 

    $this->db->query("SELECT writing_quests.id, writing_quests.points_availible, writing_quests.name, writing_quests.note, writing_quests.instructions, writing_quests.time_limit, members_quests_completed.location_completed, members_quests_completed.status FROM writing_quests LEFT OUTER JOIN members_quests_completed ON members_quests_completed.quest_id = writing_quests.id WHERE writing_quests.level_required <= '$userlevel' AND writing_quests.unlocked = 1 AND (members_quests_completed.user_id = '$user_id' OR members_quests_completed.user_id IS NULL)"); 
    $query = $this->db->get(); 
    $this->db->last_query(); 
    return $query->result(); 
}

来源

2011-01-29 JoeM05

也许CI的`查询（）`方法不喜欢换行。 – BoltClock 2011-01-29 19:48:54

感谢BoltClock。我已经无意中尝试过了。我也有用换行符编写的查询。 – JoeM05 2011-01-29 19:51:46

 
SELECT writing_quests.*, (put needed fields from members_quests_completed) 
FROM writing_quests 
LEFT JOIN members_quests_completed ....

相同的字段名在结果物体破坏结果列的名称

仅供参考它不是明智的做法*使用，你应该有字段列表...
，什么是最重要的：

 
$result = $this->db->query (...);

$这个 - > DB->查询（...）是好当INSERT或UPDATE（大部分），当你拉动的数据分配结果变量：http://codeigniter.com/user_guide/database/results.html

来源

2011-01-29 19:58:48 bensiu

添加了一个字段列表，结果相同，错误相同。这在mysql命令行中工作正常。 – JoeM05 2011-01-29 20:18:32

我想应该是这样的：

$query = $this->db->query(" 
SELECT * 
FROM writing_quests 
LEFT OUTER JOIN members_quests_completed 
ON members_quests_completed.quest_id = writing_quests.id 
WHERE writing_quests.level_required <= $userlevel 
AND ( 
    members_quests_completed.user_id = $user_id 
    OR 
    members_quests_completed.user_id IS NULL)" 
);

或者：

$query = $this->db->get('mytable');

来源

2011-01-29 20:00:36 Ivan

笨MySQL查询不工作

回答

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