Question

这个查询对我来说很复杂。第一种情况有效。我试图添加第二个案例;但是，无法让它发挥作用。一个案例在陈述时是否需要超过1个？我完全错了第二种情况吗？

我需要第二种情况的唯一原因是因为当table1.data_lock = 0时，没有table2.id等于0.我只希望它包含table2 IF table1.data_lock != 0

可能有更好的方法。

SELECT table1.id, table1.draft_sectionid, table1.page_columnid, table1.sortid, 
(case 
when table1.data_lock = 0 then table1.data 
when table1.data_lock != 0 then table2.data
end) as data,
table1.style
 FROM table1

 (case 
when table1.data_lock != 0 then JOIN table2 ON table1.data_lock=table2.id
end)

  WHERE table1.draft_sectionid = " . $section['id'] . " ORDER BY table1.sortid ASC LIMIT " . $section['columns']) or die ('Unable to execute query. '. mysqli_error($connection));

Answer 1

无法使用case-when来定义select的结构。

但是，在这种情况下，您可以使用subselect生成值：

SELECT table1.id, table1.draft_sectionid, table1.page_columnid, table1.sortid, 
(case 
when table1.data_lock = 0 then table1.data 
when table1.data_lock != 0 then (select table2.data from table2 where table2.id = table1.data_lock)
end) as data,
table1.style
 FROM table1
WHERE ...

Answer 2

不，您可以尝试使用CASE表达式来更改连接表。您可以使用外部联接

将查询更改为如下所示

SELECT table1.id, table1.draft_sectionid, table1.page_columnid, table1.sortid, 
(case when table1.data_lock = 0 then table1.data 
      else table2.data end) as data,
table1.style

FROM table1
LEFT JOIN table2 ON table1.data_lock != 0 AND table1.data_lock = table2.id
WHERE table2.id IS NOT NULL 
AND table1.draft_sectionid = " . $section['id'] . " 
ORDER BY table1.sortid ASC 
LIMIT " 1;

Answer 3

SELECT table1.id, table1.draft_sectionid, table1.page_columnid, table1.sortid, 
(case 
when table1.data_lock = 0 then table1.data 
when table1.data_lock != 0 then table2.data
end) as data,
table1.style
 FROM table1 JOIN table2 ON table1.data_lock=table2.id and table1.data_lock != 0 

WHERE table1.draft_sectionid = " . $section['id'] . " ORDER BY table1.sortid ASC LIMIT " . $section['columns']) or die ('Unable to execute query. '. mysqli_error($connection));
mysqli

使用此查询。

你能在MySQLi查询中使用多个案例吗？

3 个答案: