为了得到 animal_name 你可以使用条件聚合

SELECT animal_name
FROM table1
GROUP BY animal_name
HAVING SUM(CASE WHEN tag_name = 'cat' AND tag_group = 'species' THEN 1 ELSE 0 END  ) > 0
AND SUM(CASE WHEN tag_name = 'blue' AND tag_group = 'colour' THEN 1 ELSE 0 END  ) > 0
;

使用mysql，您还可以使用 sum(a=b) , SUM(tag_name = 'blue' AND tag_group = 'colour') 作为速记，在里面使用表达式 sum() 将产生布尔值0/1，真/假
演示
为了蓝狗

SELECT animal_name
FROM table1
GROUP BY animal_name
HAVING SUM(CASE WHEN tag_name = 'dog' AND tag_group = 'species' THEN 1 ELSE 0 END  ) > 0
AND SUM(CASE WHEN tag_name = 'blue' AND tag_group = 'colour' THEN 1 ELSE 0 END  ) > 0
;

根据最后的评论，我需要蓝猫和蓝狗加上 OR 有两个动物名称的标准

SELECT animal_name
FROM table1
GROUP BY animal_name
HAVING (
  SUM(CASE WHEN tag_name = 'dog' AND tag_group = 'species' THEN 1 ELSE 0 END  ) > 0
  AND 
  SUM(CASE WHEN tag_name = 'blue' AND tag_group = 'colour' THEN 1 ELSE 0 END  ) > 0
) OR (
  SUM(CASE WHEN tag_name = 'cat' AND tag_group = 'species' THEN 1 ELSE 0 END  ) > 0
  AND
  SUM(CASE WHEN tag_name = 'blue' AND tag_group = 'colour' THEN 1 ELSE 0 END  ) > 0
)

演示

您可以将子请求与您的条件一起使用：

SELECT * FROM Table 
INNER JOIN (
SELECT animal_name, sum(tag_name = 'blue' OR tag_name = 'cat') as sum_condition
FROM TABLE
GROUP BY animal_name
HAVING sum_condition = 2) datas USING(animal_name)

编辑以回复评论

SELECT * FROM Table 
INNER JOIN (
SELECT animal_name, sum(tag_name = 'blue' OR (tag_name IN ('cat','dog'))) as sum_condition
FROM TABLE
GROUP BY animal_name
HAVING sum_condition = 2) datas USING(animal_name)

编辑：我的错我忘了一个括号http://sqlfiddle.com/#!9/e9e941/15/0