Php 坚持用sum选择不同的记录

对不起，伙计们，我试过了，但我不得不问。我带着这个哪儿也不去

Here's my table `likes`:
"id"    "type"  "parent" "country" "votes"
"1"          "1"    "0" "US"    NULL
"2"          "2"    "1" "US"    NULL
"3"          "3"    "2" "US"    NULL
"4"          "10"   "3" "US"    "5"
"5"          "10"   "3" "US"    "15"
"6"          "10"   "3" "US"    "25"
"7"          "3"    "2" "US"    NULL
"8"          "10"   "7" "US"    "40"
"9"          "10"   "7" "US"    "25"
"10"     "10"   "7" "US"    "60"
"19"     "3"    "1" "US"    NULL
"20"     "10"   "19"    "US"    "10"
"21"     "10"   "19"    "US"    "20"
"22"     "10"   "19"    "US"    "30"

由此，我尝试选择不同的父项，类型=10，国家='us'的总和投票

我试过这样做：选择SUMlikes.vows FROM SELECT DISTINCT parent，vows FROM likes，type=10 likes；但我最终得到了非常奇怪的结果

我该怎么做

我试图得到的结果是：

parent | total
3      | 45
7      | 125
9      | 60

试着这样做

SELECT parent,sum(votes) as total FROM `likes` 
WHERE type = 10 GROUP BY parent ORDER BY parent ASC

检查

---

伊加兹！谁会想到这么简单？@Gautham3164是的。谢谢请稍等。

SELECT parent,sum(votes) as total FROM `likes` 
WHERE type = 10 GROUP BY parent ORDER BY parent ASC

select distinct parent, sum(votes) from likes 
where type = 10 and country = 'us' GROUP BY parent;