php - 拉拉维尔 5.4 : Converting a raw SQL query in Laravel Eloquent

Question

我正在尝试重写此 SQL 查询，但我陷入了困境

该查询旨在通过使用子查询仅连接最新条目来将项目表连接到project_progress表

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

致:

$projects = App\Project::where('member_id', 1)
    ->join('project_progress', function ($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();

我认为这行有问题，但我不知道如何写

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

ErrorException (E_ERROR) strtolower() 期望参数 1 为字符串，给定的对象为\vendor\laravel\framework\src\Illuminate\Database\Grammar.php

Answer 1

使用where():

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