php - 内连接-Laravel

Question

我的数据库中有三个表。用户、雇主和工作。

一些用户是发布了一些工作的雇主。

我正在尝试按用户显示作业。我的代码：用户模型

public function jobs(){
        return $this->hasManyThrough('App\Employer','App\Job');
    }

路线：

Route::get('/find_user_jobs',function(){
    $user=User::find(1);
    foreach($user->jobs as $job){
        echo $job->created_at."<br>";
    }
});

但我得到这个错误

 Column not found: 1054 Unknown column 'jobs.user_id' in 'field list' (SQL: select `employers`.*, `jobs`.`user_id` from `employers` inner join `jobs` on `jobs`.`id` = `employers`.`job_id` where `jobs`.`user_id` = 1)

我知道它试图在工作中找到 user_id 但这就是我想要它做的

我对程序的渴望 当我给用户 id 时，去雇主表搜索 user_id，如果它存在，去工作表并搜索雇主 ID 并返回所有具有雇主 ID 的工作。

用户迁移

<?php

use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;

class CreateUsersTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('users', function (Blueprint $table) {
            $table->increments('id');
            $table->string('name');
            $table->string('email')->unique();
            $table->string('password');
            $table->integer('employer_id');
            $table->rememberToken();
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::drop('users');
    }
}

工作迁移

<?php

use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;

class CreateJobsTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('jobs', function (Blueprint $table) {
            $table->increments('id');
            $table->integer('employer_id');
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::drop('jobs');
    }
}

雇主移民

<?php

use Illuminate\Database\Schema\Blueprint;
use Illuminate\Database\Migrations\Migration;

class CreateEmployersTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {
        Schema::create('employers', function (Blueprint $table) {
            $table->increments('id');
            $table->string('company_name');
            $table->integer('user_id');
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::drop('employers');
    }
}

Answer 1

因为谁有Employee你可以创建一个扩展模型的模型UserJobsApp\EmployeeApp\User

class Job {

    public function employee()
    {
        return $this->belongsTo(App\Employee::class,'user_id');
    }
}

并像这样创建一个 Employee 类

在Employee模型中，我将$table属性设置为users，当我们进行一些查询时，该查询将设置目标表users而不是employees表，这将是 Eloquent 的默认行为。

class Employee extends User
{
    protected $table = "users";

    public function jobs()
    {
        return $this->hasMany(Job::class, 'user_id');
    }
}

您可以直接使用Employee模型并获取jobs

这是相应的迁移

create_user_table

class CreateUsersTable extends Migration
{

    public function up()
    {
        Schema::create('users', function (Blueprint $table) {
            $table->increments('id');
            $table->string('name');
            $table->string('email')->unique();
            $table->string('password');
            $table->string('company_name')->nullable();
            $table->rememberToken();
            $table->timestamps();
        });
    }

    public function down()
    {
        Schema::drop('users');
    }
}

create_job_table

class CreateJobsTable extends Migration
{

    public function up()
    {
        Schema::create('jobs', function (Blueprint $table) {
            $table->increments('id');
            $table->integer("user_id")->unsigned();
            $table->foreign('user_id')->references('id')->on('users');
            $table->timestamps();
        });
    }

    public function down()
    {
        Schema::drop('jobs');
    }
}

Answer 2

关系参数的顺序错误：

public function jobs(){
    return $this->hasManyThrough('App\Job', 'App\Employer');
}