如何从另一个表中通过他们的ID从另一个表中获取值？

Question

文件表（req1-7是需求的ID从另一个表）

| document_id | document_name | document_amount | req_1 | req_2 | req_3 | req_4 | req_5 | req_6 | req_7 | 
|  1  | Diploma  |  40   | 1 | 3 | 4 |  4 | 6 | 7 | 8 | 

要求表

| requirement_id | requirement_name | 
| 1   | 1 x 1 picture | 
| 2   | 2 x 2 picture | 
| 3   | Registration form| 
| 4   | Clearance  | 
| 5   | Medical cert  | 
| 6   | xray result  | 
| 7   | excuse letter | 
| 8   | affidavit  | 
| 9   | comsoc clearance | 

预期的结果是类似的文件表，但与相应的显示的要求了名他们身份证的价值。

我想知道正确的语法来查询这些

到目前为止，我的查询是：

SELECT * FROM document_tbl 
      WHERE requirement1,requirement2,requirement3,requirement4,requirement5,requirement6,requirement7 
      IN (
       SELECT requirement_name FROM requirements_tbl WHERE requirement_id=requirement1,requirement2,requirement3,requirement4,requirement5,requirement6,requirement7)"; 

但我screwedup莫名其妙。先谢谢您的帮助。我会很感激。

Answer 1

像这样：

SELECT d.document_id, 
    d.document_name, 
    d.document_amount, 
    r1.requirement_name AS req_1, 
    r2.requirement_name AS req_2, 
    r3.requirement_name AS req_3, 
    r4.requirement_name AS req_4, 
    r5.requirement_name AS req_5, 
    r6.requirement_name AS req_6, 
    r7.requirement_name AS req_7 
FROM documents AS d, 
    requirements AS r1, 
    requirements AS r2, 
    requirements AS r3, 
    requirements AS r4, 
    requirements AS r5, 
    requirements AS r6, 
    requirements AS r7 
WHERE d.req_1 = r1.requirement_id 
    AND d.req_2 = r2.requirement_id 
    AND d.req_3 = r3.requirement_id 
    AND d.req_4 = r4.requirement_id 
    AND d.req_5 = r5.requirement_id 
    AND d.req_6 = r6.requirement_id 
    AND d.req_7 = r7.requirement_id 

等。

Answer 2

中间体表（document_requirements）

| document_id | requirement_id | 
|  1  | 1    | 
|  1  | 3    | 
|  1  | 4    | 
|  1  | 6    | 
|  1  | 6    | 
|  1  | 7    | 
|  1  | 8    | 

查询

SELECT d.document_id,dr.requirement_id,r.requirement_name 
FROM documents AS d 
JOIN document_requirements AS dr ON (dr.document_id=d.document_id) 
JOIN requirements AS r ON (r.requirement_id=dr.requirement_id) 
WHERE d.document_id = 1;