table2 是外部连接但条件 table2.user_id IN (...) 在where子句中，将查询更改为内部联接。将条件从 WHERE 至 ON 条款：

select t1.name, t1.id, count(t2.id) as count
from table1 as t1
left join table2 as t2 on
  find_in_set(t1.id, t2.table1_ids) > 0 and
  t2.user_id in (1, 2)
group by t1.name, t1.id

sql小提琴
附言： WHERE 1 IN ('1,2') 尝试转换 '1,2' 一个数字，因此匹配1。

我可能会因此恨我自己，但这可能有用：

select
    t1.name,
    t1.id,
    count(t2.id) as count
from
    table1 as t1
left join
    table2 as t2 on
        (
            -- We need to account for all of the variations of finding t1.id in the comma-separated field
            t2.table1_ids = t1.id or -- exactly this ID
            t2.table1_ids LIKE concat( t1.id, ',%' ) or -- starts with this ID
            t2.table1_ids LIKE concat( '%,', t1.id ) or -- ends with this ID
            t2.table1_ids LIKE concat( '%,', t1.id, ',%' ) -- the ID is found between two commas
        )
where
    t2.user_id in (1,2)
group By
    t1.name, t1.id

正如一位评论员所建议的，我们应该避免在表2中添加逗号分隔的数据，因为这是一种不好的做法。
不过，也就是说，您可以使用laravel的查询生成器来构建您的查询，使其更可读、更清晰。基于salman a关于更改where to on的观点，您可以这样做：

DB::table("table1 as t1")
    ->leftJoin("table2 as t2", function($join) {
        $join->on(\DB::raw("find_in_set(t1.id, t2.table1_ids) > 0 and t2.user_id in (1, 2)"), \DB::raw(""), \DB::raw(""));
    })
    ->select("t1.name", "t1.id", \DB::raw("count(t2.id) as count"))
    ->groupBy("t1.name", "t1.id")
    ->get();