Question

我有多对多关系的 students 和 subjects 表（数据透视表是 student_subject）。

学生模型

public function subjects()
{
    return $this->belongsToMany(Subject::class, 'student_subject');
}

学科模型

public function students()
{
    return $this->belongsToMany(Student::class, 'student_subject');
}

这里我想要特定的学生科目计数。我尝试了以下方法，效果很好，但我想要最有效的方法。

1.

$student = Student::find($id);
$subject_count = $student->subjects()->count();

我通过 laravel 调试器检查了 SQL 查询，它显示如下

select * from `students` where `students`.`id` = '10' limit 1

select count(*) as aggregate from `subjects` inner join `student_subject` on `subjects`.`id` = `student_subject`.`subject_id` where `student_subject`.`student_id` = 10 and `subjects`.`deleted_at` is null

$student = Student::withCount('subjects')->find($id);
$subject_count = $student->subjects_count;

我通过 laravel 调试器检查了 SQL 查询，它显示如下

select `students`.*, (select count(*) from `subjects` inner join `student_subject` on `subjects`.`id` = `student_subject`.`subject_id` where `students`.`id` = `student_subject`.`student_id` and `subjects`.`deleted_at` is null) as `subjects_count` from `students` where `students`.`id` = '10' limit 1

$student = Student::find($id);
$subject_count = $student->loadCount('subjects')->subjects_count;

我通过 laravel 调试器检查了 SQL 查询，它显示如下

select * from `students` where `students`.`id` = '10' limit 1

select `id`, (select count(*) from `subjects` inner join `student_subject` on `subjects`.`id` = `student_subject`.`subject_id` where `students`.`id` = `student_subject`.`student_id` and `subjects`.`deleted_at` is null) as `subjects_count` from `students` where `students`.`id` in (10)

$student = Student::find($id);
$subject_count = DB::table('student_subject')->where('student_id', $student->id)->count();

我通过 laravel 调试器检查了 SQL 查询，它显示如下

select * from `students` where `students`.`id` = '10' limit 1

select count(*) as aggregate from `student_subject` where `student_id` = 10

根据上面的方法哪一种最好，为什么？或者是否还有其他不同的最佳方式？

Answer 1

执行 relation()->count() 可能会更快。

但如果您只需要计数，那么 withCount() 在内存消耗方面应该会更好。

在laravel中获得多对多关系计数的最佳有效方法是什么？

1 个答案: