Question

代码

   function viewPDF()
{
    $reports = Report::join('president_report', 'reports.id', '=', 'president_report.report_id')
        ->join('president_report as pr', 'presidents.id', '=', 'pr.president_id')
        ->join('quarter_report', 'president_report.report_id', '=', 'quarter_report.report_id')
        ->filter()->latest()->get();
    $pdf = PDF::loadView('reports.test1', ['reports' => $reports]);
    return $pdf->stream('reports.pdf');
}

错误

SQLSTATE[42S22]: Column not found: 1054 Unknown column 'presidents.id' in 'on clause' (SQL: select * from `reports` inner join `president_report` on `reports`.`id` = `president_report`.`report_id` inner join `president_report` as `pr` on `presidents`.`id` = `pr`.`president_id` inner join `quarter_report` on `president_report`.`report_id` = `quarter_report`.`report_id` order by `created_at` desc)

我想用口才来做。拜托，我该怎么办？

1-表格 quarters

Schema::create('quarters', function (Blueprint $table) {
        $table->increments('id');
        $table->string('title');
        $table->timestamps();
    });
    Schema::create('quarter_report', function (Blueprint $table) {
        $table->integer('quarter_id')->unsigned();
        $table->integer('report_id')->unsigned();
        $table->primary(['quarter_id' , 'report_id']);
    });

2-表格 presidents

Schema::create('presidents', function (Blueprint $table) {
        $table->Increments('id');
        $table->string('P_name');
        $table->timestamps();
    });
 Schema::create('president_report', function (Blueprint $table) {
        $table->integer('president_id')->unsigned();
        $table->integer('report_id')->unsigned();
        $table->primary(['president_id','report_id']);
    });

3-表格 reports

Schema::create('reports', function (Blueprint $table) {
        $table->Increments('id');
        $table->integer('user_id')->unsigned();
        $table->string('target_value');
        $table->text('major_activity');
        $table->string('indicator');
        $table->string('responsibility');
        $table->string('progress_activity');
        $table->string('documents')->nullable();
        $table->string('percentage_activity')->nullable();
        $table->text('problem')->nullable();
        $table->text('solution')->nullable();
        $table->text('description')->nullable();
        $table->string('image')->default('default.png');
        $table->tinyInteger('status')->default('0');
        $table->tinyInteger('active')->default('0');
        $table->timestamps();
    });

I have 5 tables

Answer 1

这是this version of question的答案。

尝试使用别名加入表，如下所示

->join('president_report as pr', 'presidents.id', '=', 'pr.president_id')

和/或改为'pr.president_id'使用DB::raw( 'pr.president_id' )

Answer 2

使用DB :: raw（）

function viewPDF()
{
    $reports = Report::join('president_report','president_report.report_id' 'reports.id', '=', 'president_report.report_id')
        ->join('president_report as pr', DB::raw('presidents.id'), '=',DB::raw( 'pr.president_id'))
        ->join('quarter_report', 'president_report.report_id', '=', 'quarter_report.report_id')
        ->filter()->latest()->get();
    $pdf = PDF::loadView('reports.test1', ['reports' => $reports]);
    return $pdf->stream('reports.pdf');
}

Answer 3

我认为您已与模型

建立了联系

（1）第一个选项

CONTROLLER

在您的控制器中，您可以这样做。

President::with('president_report')->get()->map(function($president){
  return [
    'president_id' => $president->id,
    'president_report_id' => $president->president_report->id,
  ]
})

（2）第二选项

如果要访问父级父级关系。

quarters属于president_report，而president_reports属于president

您可以这样做。

总裁模式

public function quarters() {
  return $this->hasManyThrough(Quarter::class, President_report::class);
}

CONTROLLER

您可以在控制器中执行此操作。

$presidents = President::all();

$quaters = president->quarters()->where('quarters.report_id', 1)->get();

dd($quarters);

如何使用laravel雄辩地联接多个表？

3 个答案: