Php 选择带有内部联接的计数

我有两张桌子：

project (id, name)
list (id, name, project_id)

我想把每一个项目和清单的数量都加入到这个项目中

我试过：

SELECT project.id, project.name, COUNT(list.id) AS list_count
FROM project 
INNER JOIN list ON (list.project_id = project.id)
GROUP BY project.id DESC

但我只得到有列表的项目

如果项目没有自己的列表，我就不明白

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如果count（list\u id）=0，我想要list\u count=0，我必须怎么做？请使用左连接而不是内部连接

SELECT project.id, project.name, COUNT(list.id) AS list_count
FROM project 
Left JOIN list ON (list.project_id = project.id)
GROUP BY project.id DESC

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使用左联接而不是内联接

SELECT project.id, project.name, COUNT(list.id) AS list_count
FROM project 
Left JOIN list ON (list.project_id = project.id)
GROUP BY project.id DESC