0

我有 3 张桌子：

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data_title从一个tb_wrapper_scho结果中tb_wrapper_li得到相同doc_name的结果document_titletb_document$title

这是查询：

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);
}

它给了我空白的结果。请帮帮我..谢谢:)

4

5 回答 5

0

由于您的表包含具有相同名称的列，因此在使用别名检索数据时，您必须将列名称更改为不同的名称。因此结果数组必须包含每个值的唯一索引。所以你的查询必须是

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'
于 2012-11-01T13:40:24.640 回答
0 
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);

有什么好处吗？

您还可以检查 mysql_query 的结果，并且 mysql_fetch_array() 应该返回 TRUE。

于 2012-11-01T13:42:49.003 回答
0

您缺少该子句tb_document中的表格。FROM试试这个：

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'
于 2012-11-01T12:54:35.760 回答
0

我有想法，你喜欢吗：

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

仅此而已，祝朋友好运！

于 2012-11-01T13:30:10.453 回答
0 
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'
于 2012-11-01T12:57:28.577 回答