mysql选择。如果WHERE中的值不存在，则根据WHERE中的另一个（值）选择

我有这样的桌子

我有两个变量/值，例如

FLASH

和

LCD

我想要：如果

name

列中不存在

FLASH

，则

选择。。。其中

column

name='LCD'

。如果存在

FLASH

，则不基于

LCD

我写了这样的疑问

SELECT 

t1.name AS lev1, t1.url AS url1

FROM CategoryForTest AS t1
LEFT JOIN CategoryForTest AS t2 ON t2.parent = t1.category_id
LEFT JOIN CategoryForTest AS t3 ON t3.parent = t2.category_id
LEFT JOIN CategoryForTest AS t4 ON t4.parent = t3.category_id

WHERE exists (
SELECT name
from CategoryForTest
WHERE t4.name = 'FLASH'
)

UNION ALL

SELECT 

t1.name AS lev1, t1.url AS url1

FROM CategoryForTest AS t1
LEFT JOIN CategoryForTest AS t2 ON t2.parent = t1.category_id
LEFT JOIN CategoryForTest AS t3 ON t3.parent = t2.category_id
LEFT JOIN CategoryForTest AS t4 ON t4.parent = t3.category_id

WHERE not exists (
SELECT name
from CategoryForTest
WHERE t4.name = 'FLASH'
)
AND t2.name = 'LCD'

但我明白了

Array
(
[0] => Array
    (
        [lev1] => ELECTRONICS
        [url1] => url-electronics
    )

[1] => Array
    (
        [lev1] => TELEVISIONS
        [url1] => url-televisions
    )

)

---

希望只看到

[0]=>数组

基于此找到解决方案

如果

FLASH

存在，则基于

FLASH

获取数据。如果

FLASH

不存在，则基于

LCD

进行提取，这可能会对您有所帮助

(
   SELECT 

     t1.name AS lev1, t1.url AS url1

   FROM test AS t1
     LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
     LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
     LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

   WHERE not exists (
       SELECT 

         t1.name AS lev1, t1.url AS url1

       FROM test AS t1
         LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
         LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
         LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

      WHERE exists (
         SELECT name
         from test
         WHERE t4.name = 'FLASH'
      )
   )
   AND t2.name = 'LCD'
 GROUP BY t1.categ_id
)
UNION ALL
(
   SELECT 

     t1.name AS lev1, t1.url AS url1

   FROM test AS t1
     LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
     LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
     LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

   WHERE exists (
     SELECT name
     from test
     WHERE t4.name = 'FLASH'
   )
)

---

fidle is

如何使用两个单独的查询，首先用于获取名称为FLASH的查询，然后如果未找到任何结果，则选择名称为LCD的查询。两个单独的查询将使用额外的资源？不确定它对性能的影响有多大（比较两个或一个查询）。我可能还需要一个查询来从另一个表获取产品。因此，只有当我将

FLASH

更改为

FLASH1

时，才会出现3个查询。我使用

LCD

@user2118559获得一个长列表，您想忽略1？我试图更改为

FLASH1

（mysql中不存在的值，用于执行基于

LCD

的查询）并查看多个

LCD

，但只需要查看一个（与mysql中一样，只有一个

LCD

）@user2118559表中是否有任何唯一字段可用于group by（）？因此，请在group by中使用category_id。

(
   SELECT 

     t1.name AS lev1, t1.url AS url1

   FROM test AS t1
     LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
     LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
     LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

   WHERE not exists (
       SELECT 

         t1.name AS lev1, t1.url AS url1

       FROM test AS t1
         LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
         LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
         LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

      WHERE exists (
         SELECT name
         from test
         WHERE t4.name = 'FLASH'
      )
   )
   AND t2.name = 'LCD'
 GROUP BY t1.categ_id
)
UNION ALL
(
   SELECT 

     t1.name AS lev1, t1.url AS url1

   FROM test AS t1
     LEFT JOIN test AS t2 ON t2.parent = t1.categ_id
     LEFT JOIN test AS t3 ON t3.parent = t2.categ_id
     LEFT JOIN test AS t4 ON t4.parent = t3.categ_id

   WHERE exists (
     SELECT name
     from test
     WHERE t4.name = 'FLASH'
   )
)