$data = $con->query("SELECT id, content FROM table1 WHERE id IN (". $column['data'] .")");


我上面有需要添加if then语句的代码。

if table1.draftid = 0, then select content from table 1
if table1.draftid != 0, then select content from table 2



  仅供参考...。
  如果TABLE1.DRAFTID!= 0,然后table1.draftid = table2.id


我尝试了以下方法。

$data = $con->query("SELECT table1.id,

(case
when table1.draftid = 0 then table1.content
when table1.draftid != 0 then (select table2.content from table2 where table1.draftid = table2.id)
end) as data,


FROM table1
JOIN table2 ON table1.draftid = table2.id
WHERE table1.id IN (". $column['data'] .")");


使用例

> Table 1
>
>id = 1
>
>draftid = 1
>
>content = Table 1 Test

---------

> Table 2
>
> id = 1
>
> content = Table 2 Content





  所需的结果:If table 1 draftid = 0,则它将调用
  表1中的内容。但是,如果table1.draftid != 0(例如:1),则>它将从表2中的内容中提取table1.draftid = table2.id

最佳答案

你可以试试这个吗?

$data = $con->query("SELECT table1.id,

(case
when table1.draftid = 0 then table1.content
when table1.draftid != 0 then  table2.content
end) as data,


FROM table1
JOIN table2 ON table1.draftid = table2.id
WHERE table1.id IN (". $column['data'] .")");

关于mysql - 如果/然后陈述,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/35409195/

10-17 03:06