Php Laravel5.4：在LaravelEloquent中转换原始SQL查询

我正试图重写这个SQL查询，但我在这一点上卡住了

该查询旨在通过使用子查询将projects表连接到project_progress表，以仅连接最新条目

SELECT * FROM projects
JOIN project_progress ON project_progress.id = 
(
    SELECT id FROM project_progress
    WHERE project_progress.project_id = projects.id
    ORDER BY project_progress.created_at DESC
    LIMIT 1
)
WHERE project_progress.next_action_date < NOW()
AND projects.status != 'Complete'
AND projects.member_id = 1
ORDER BY projects.title ASC

ErrorException（E_ERROR）strtolower（）要求参数1为字符串，对象为给定的\vendor\larvel\framework\src\light\Database\Grammar.php

---

“我觉得这句话有点不对劲”你为什么会出错？还是返回了错误的结果？您可以使用

->toSql（）

函数检查生成的SQLErrorException（E_ERROR）strtolower（）期望参数1为字符串，object-given\vendor\laravel\framework\src\illighte\Database\Grammar.php

$projects = App\Project::where('member_id', 1)
    ->join('project_progress', function ($join) {
        $join->on('project_progress.id', '=', function ($query) {
            $query->select('project_progress.id')
                ->from('project_progress')
                ->where('project_progress.project_id', 'projects.id')
                ->orderBy('project_progress.created_at', 'desc')
                ->limit(1);
        });
    })
    ->where('project_progress.next_action_date', '<', Carbon\Carbon::now())
    ->notCompleted()
    ->orderBy('projects.project_title', 'asc')
    ->get();

$join->on('project_progress.id', '=', function ($query) {

$join->where('project_progress.id', '=', function ($query) {