我正在尝试重写此SQL查询,但此时此刻我被困住了
该查询旨在通过使用子查询仅将最新条目连接到Projects表到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
至:
$projects = App\Project::where('member_id', 1)
->join('project_progress', function ($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();
我认为此行有些错误,但是我不确定如何写
$join->on('project_progress.id', '=', function ($query) {
ErrorException(E_ERROR)strtolower()期望参数1为字符串,对象指定为\ vendor \ laravel \ framework \ src \ Illuminate \ Database \ Grammar.php
最佳答案
使用where()
:
$join->where('project_progress.id', '=', function ($query) {