Php MYSQL关联计数

所以我试图计算一些评论，当我显示公司时，我想在每个公司旁边显示评论计数，比如在博客上，当你在帖子列表上显示评论总数时

我的问题是：

SELECT COUNT(*) as total FROM `company_admin_notes`
WHERE company_id IN(49152,49408,49155) GROUP by company_id;

它给了我这样的东西：

所以我需要像这样的东西

 total | company_id
 5     | 49152
 4     | 4908

因此，我知道哪家公司有哪一笔资金，或者如果你们有任何其他想法，我如何才能实现这样的目标。

试试这个：

SELECT COUNT(*) as total, company_id FROM `company_admin_notes`WHERE company_id IN(49152,49408,49155) GROUP by company_id;

SELECT COUNT(*) as total,company_id FROM `company_admin_notes`
WHERE company_id IN(49152,49408,49155) GROUP by company_id
HAVING COUNT(*)>0;

试试这个：

SELECT COUNT(*) as total, company_id FROM `company_admin_notes`WHERE company_id IN(49152,49408,49155) GROUP by company_id;

SELECT COUNT(*) as total,company_id FROM `company_admin_notes`
WHERE company_id IN(49152,49408,49155) GROUP by company_id
HAVING COUNT(*)>0;

试试这个：

SELECT COUNT(*) as total, company_id FROM `company_admin_notes`WHERE company_id IN(49152,49408,49155) GROUP by company_id;

SELECT COUNT(*) as total,company_id FROM `company_admin_notes`
WHERE company_id IN(49152,49408,49155) GROUP by company_id
HAVING COUNT(*)>0;

试试这个：

SELECT COUNT(*) as total, company_id FROM `company_admin_notes`WHERE company_id IN(49152,49408,49155) GROUP by company_id;

SELECT COUNT(*) as total,company_id FROM `company_admin_notes`
WHERE company_id IN(49152,49408,49155) GROUP by company_id
HAVING COUNT(*)>0;

---

在我看来是错误的，您有

COUNT（*），company\u id作为总计

，但它应该是

COUNT（*）作为总计，company\u id

。。。对吧？哇，太棒了！我会尽快接受的！非常感谢。也许如果你也可以添加一个评论什么是different@naththedeveloper哎呀，谢谢你指出这一点！！Fixed Now在我看来是错误的，您有

COUNT（*），company\u id作为总计

，但它应该是

COUNT（*）作为总计，company\u id

。。。对吧？哇，太棒了！我会尽快接受的！非常感谢。也许如果你也可以添加一个评论什么是different@naththedeveloper哎呀，谢谢你指出这一点！！现在修好了