2014-08-27 145 views
0

场景:
我需要选择20个用户与他们的所有信息,还可以选择与他们所教的语文老师的个人资料。我试图做到这一点,但似乎MySQL只返回一个值的数组没有嵌套的数组值,有没有办法做到这一点?MySQL的 - 从表中选择所有从另一个表中选择相关行

表:

+------------------------------------------+ 
|     users     | 
+------------------------------------------+ 
| id | firstname |  email  | 
+------------------------------------------+ 
| 18 |  Tom  | [email protected] | 
+------------------------------------------+ 
| 30 | Jerry | [email protected]  | 
+------------------------------------------+ 
| 25 | Butch | [email protected] | 
+------------------------------------------+ 

+------------------------------------+ 
|    teachers    | 
+------------------------------------+ 
| id | user_id | trial_lessons | 
+------------------------------------+ 
| 10 | 18  |  yes  | 
+------------------------------------+ 
| 26 | 30  |  no  | 
+------------------------------------+ 
| 28 | 25  |  no  | 
+------------------------------------+ 

+------------------------------------------+ 
|   teacher_languages    | 
+------------------------------------------+ 
| id | teacher_id | language_text_id | 
+------------------------------------------+ 
| 16 |  10  |   6   | 
+------------------------------------------+ 
| 40 |  10  |   8   | 
+------------------------------------------+ 
| 16 |  28  |   6   | 
+------------------------------------------+ 
| 16 |  28  |  10   | 
+------------------------------------------+ 
| 16 |  26  |   6   | 
+------------------------------------------+ 

+-------------------+ 
|  languages  | 
+-------------------+ 
| id | language | 
+-------------------+ 
| 6 | English | 
+-------------------+ 
| 8 | French | 
+-------------------+ 
| 10 | Spanish | 
+-------------------+ 

到目前为止我的代码

SELECT 
    users.*,   
    nationality.country AS country_of_nationality, 
    residence.country AS country_of_residence, 
FROM 
    users 
LEFT JOIN 
    text_countries AS nationality 
ON 
    users.nationality = nationality.iso_code_2 
AND 
    nationality.language_id = ? 
LEFT JOIN 
    text_countries AS residence 
ON 
    users.residence_country = residence.iso_code_2 
AND 
    residence.language_id = ? 
ORDER BY 
    users.created_at 
DESC LIMIT 
     20 

预期成果

[0] => 
    [user_id] => 18 
    [firstname] => 'Tom' 
    [teacher_id] => 10 
    [languages] => 
       [language] => 'English' 
       [language] => 'French' 
[1] => 
    [user_id] => 30 
    [firstname] => 'Jerry' 
    [teacher_id] => 26 
    [languages] => 
       [language] => 'English' 
[2] => 
    [user_id] => 25 
    [firstname] => 'Butch' 
    [teacher_id] => 28 
    [languages] => 
       [language] => 'English' 
       [language] => 'Spanish' 
+0

向我们展示您的代码,请在寻求帮助之前亲自尝试一下。 – Adam 2014-08-27 22:40:21

+0

MySQL不返回多维记录。每个记录都是一维的。大多数关系数据库都是这样做的。您可以使用'GROUP_CONCAT'将多个值收集到1列中,具体取决于您需要的语言数据。 – Rudie 2014-08-27 22:41:06

+0

我已经做了,我迄今为止的代码非常长,因为它从多个表计算,如果它大部分是代码,它不会让我发布问题。干杯! – CupOfJoe 2014-08-27 22:41:39

回答

0

的MySQL没有按” t返回多维记录。每个记录都是一维的。大多数关系数据库都是这样做的。

你可以使用GROUP_CONCAT收集多个值到1列,虽然,这取决于你如何需要的语言数据:

SELECT stuff, GROUP_CONCAT(l.language) AS languages 
FROM users u 
JOIN teachers t ON .... 
JOIN teacher_languages tl ON .... 
JOIN languages l ON .... 
GROUP BY u.id 

我假设你知道(和有吗?)的JOIN小号...您需要一个额外的JOIN(多个)至teacher_languages,然后再使用另一个JOIN(单个)至languages,然后使用GROUP_CONCAT将多个语言记录收集为1个值。

+0

我尝试了你的建议,它很好地工作,现在的问题是它返回一个用户的信息的单个值,并连接所有可用的语言进入它。 – CupOfJoe 2014-08-27 23:23:49

+0

我错过了'GROUP BY u.id',它让我感到困惑,就像你在代码中写的那样,为什么你这样写呢? – CupOfJoe 2014-08-27 23:32:55

+0

我不知道=)我想我可以只说'GROUP BY u.id' ...我改变了答案。对不起,不太清楚。它现在工作吗? – Rudie 2014-08-28 10:22:23

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