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1個結果中3個表的內部聯接

[英]Inner join from 3 tables in one result

我有3張桌子:

tb_document
============================
|document_id|document_title|
============================
|     1     |        A     |
|     2     |        B     |
|     3     |        C     |
============================

tb_wrapper_scho                  tb_wrapper_li
=============================    =============================
|id_scho|data_title|doc_name|    |id_sli |data_title|doc_name|
=============================    =============================
|   1   |  TitleA  |   A    |    |   1   |  TitleB  |   B    |
|   2   |  TitleC  |   C    |    =============================
=============================

我想從tb_wrapper_schotb_wrapper_li獲取data_title ,其中在一個結果$titletb_document中的document_title具有相同的doc_name

這是查詢:

SELECT
      tb_document.document_title,
      tb_wrapper_scho.data_title,
      tb_wrapper_scho.doc_name,
      tb_wrapper_li.data_title,
      tb_wrapper_li.doc_name
FROM
      INNER JOIN tb_wrapper_scho ON tb_document.document_title = tb_wrapper_scho.doc_name
      INNER JOIN tb_wrapper_li ON tb_document.document_title = tb_wrapper_li.doc_name
WHERE 
      document_id = '$doc_id' 

while ($row = mysql_fetch_array($sql)) {
     $title = $row['data_title'];
     print_r($title);
}

它給我空白的結果。 請幫助我..謝謝:)

您在FROM子句中缺少tb_document表。 嘗試以下方法:

SELECT
      d.document_title,
      s.data_title,
      s.doc_name,
      i.data_title,
      i.doc_name
FROM  tb_document d
      INNER JOIN tb_wrapper_scho s ON d.document_title = s.doc_name
      INNER JOIN tb_wrapper_li i ON d.document_title =i.doc_name
WHERE 
      document_id = '$doc_id' 
SELECT
      tb_document.document_title,
      tb_wrapper_scho.data_title,
      tb_wrapper_scho.doc_name,
      tb_wrapper_li.data_title,
      tb_wrapper_li.doc_name
FROM tb_document
      left outer JOIN tb_wrapper_scho ON tb_document.document_title = tb_wrapper_scho.doc_name
      left outer JOIN tb_wrapper_li ON tb_document.document_title = tb_wrapper_li.doc_name
WHERE 
      document_id = '$doc_id' 

我有主意,您想要嗎?

SELECT * FROM
  (SELECT * FROM tb_document as tbDoc INNER JOIN tb_wrapper_scho as tbScho
  ON tbDoc.document_title = tbScho.doc_name
UNION
  SELECT * FROM tb_document as tbDoc INNER JOIN tb_wrapper_li as tblI
  ON tbDoc.document_title = tblI.doc_name)threeTableJoin ORDER BY `document_id` ASC

只是,祝你好運!

由於表包含具有相同名稱的列,因此在使用別名檢索數據時,必須將列名稱更改為不同的名稱。 因此結果數組必須包含每個值的唯一索引。 因此您的查詢必須為

SELECT 
      d.document_title as document_title,
      s.data_title as scho_data_title,
      s.doc_name as scho_doc_name,
      li.data_title as li_data_title,
      li.doc_name as li_data_name
FROM
      tb_document d
      LEFT JOIN tb_wrapper_scho s ON d.document_title = s.doc_name
      LEFT JOIN tb_wrapper_li li ON li.doc_name = d.document_title 
WHERE 
      d.document_id = '$doc_id'
SELECT
      tb_document.document_title,
      tb_wrapper_scho.data_title,
      tb_wrapper_scho.doc_name,
      tb_wrapper_li.data_title,
      tb_wrapper_li.doc_name

FROM tb_document AS tb_document

    LEFT JOIN tb_wrapper_scho AS tb_wrapper_scho ON (tb_wrapper_scho.doc_name=tb_document.document_title)
    LEFT JOIN tb_wrapper_li AS tb_wrapper_li ON (tb_wrapper_li.doc_name=tb_document.document_title)

WHERE tb_document.document_id = intval($doc_id);

有什么好處嗎

您還可以檢查mysql_query的結果,並且mysql_fetch_array()應該返回TRUE。

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