查询laravel中的关系

Question

我正在练习laravel，在查询关系时感到困惑，我有两个表（用户和角色）

在我的用户模型中，我有

use SoftDeletes;

public function role(){
    return $this->hasOne(Role::class);
}

在我的榜样中，我有

use SoftDeletes;

public function user(){
   return $this->belongsTo(User::class);
}

这是我的角色表结构

这是我的用户表格结构

如何根据用户表查询角色

Answer 1

您应该在两个模型中都更新您的关系：

用户模型

use SoftDeletes;

    public function role(){
        return $this->belongsTo(Role::class,'role_id');
    }

角色模型

use SoftDeletes;

public function user(){
   return $this->hasOne(User::class);
}

您的查询如下：

更新后的答案

$user = User::find(1);
dd($user->roles());

Answer 2

您需要更改两个模型中的关系：

select t1.* from (
SELECT column_names 
FROM folders 
UNION
SELECT column_names 
FROM folders_archive) t1
WHERE t1.column_names = what_you_want;

以及其他模型中的

function howManyHundreds(num) {
  return num / 100;
  return num % 10;
}

console.log(howManyHundreds(894))

console.log(howManyHundreds(323))

，然后将其称为：

 use SoftDeletes;

public function role(){
     return $this->hasOne(Role::class,'id','role_id');
}

尝试一下，它将起作用...

Answer 3

应用/模型/用户

public function role()
    {
        return $this->belongsTo("App\Model\Role", "role_id", "role_id");
    }

app / Model / Role

public function user()
    {
        return $this->hasOne("App\Model\Users");
    }

进一步阅读：https://laravel.com/docs/5.6/eloquent-relationships#one-to-one

Answer 4

如果Laravel认为您的关系Fatal Exception: java.lang.RuntimeException: An error occurred while executing doInBackground() at android.os.AsyncTask$3.done(AsyncTask.java:353) at java.util.concurrent.FutureTask.finishCompletion(FutureTask.java:383) at java.util.concurrent.FutureTask.setException(FutureTask.java:252) at java.util.concurrent.FutureTask.run(FutureTask.java:271) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1162) at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:636) at java.lang.Thread.run(Thread.java:764) Caused by java.lang.SecurityException: Caller no longer running, last stopped +2s213ms because: timed out while starting at android.os.Parcel.readException(Parcel.java:1943) at android.os.Parcel.readException(Parcel.java:1889) at android.app.job.IJobCallback$Stub$Proxy.dequeueWork(IJobCallback.java:191) at android.app.job.JobParameters.dequeueWork(JobParameters.java:196) at android.support.v4.app.JobIntentService$JobServiceEngineImpl.dequeueWork(SourceFile:314) at android.support.v4.app.JobIntentService.dequeueWork(SourceFile:639) at android.support.v4.app.JobIntentService$CommandProcessor.doInBackground$10299ca(SourceFile:2389) at android.os.AsyncTask$2.call(AsyncTask.java:333) at java.util.concurrent.FutureTask.run(FutureTask.java:266) at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1162) at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:636) at java.lang.Thread.run(Thread.java:764) 应该有用。

只需在代码中的任意位置尝试User::find(1)->role。

Answer 5

用户模型

use SoftDeletes;

public function role(){
    return $this->hasOne(Role::class,'role_id','id);
}

角色模型

use SoftDeletes;

public function user(){
   return $this->belongsTo(User::class,'id','role_id');
}

像这样更新您的代码，它就可以工作。