Laravel 5.6:在数据库中创建模式

时间:2018-09-21 14:59:37

标签: php laravel postgresql laravel-5

我正在使用postgresql。我正在使用以下命令创建数据库:

<?php

namespace efsystem\Console\Commands;

use Illuminate\Console\Command;

class CreatePostgressDatabase extends Command
{
/**
 * The name and signature of the console command.
 *
 * @var string
 */
protected $signature = 'pgsql:createdb {name?}';

/**
 * The console command description.
 *
 * @var string
 */
protected $description = 'Create a new pgsql database schema based on the database config file';

/**
 * Create a new command instance.
 *
 * @return void
 */
public function __construct()
{
    parent::__construct();
}

/**
 * Execute the console command.
 *
 * @return mixed
 */
public function handle()
{
    $dbname = config('database.connections.pgsql.database');
    $dbuser = config('database.connections.pgsql.username');
    $dbpass = config('database.connections.pgsql.password');
    $dbhost = config('database.connections.pgsql.host');

    try {
                $db = new \PDO("pgsql:host=$dbhost", $dbuser, $dbpass);

                $test = $db->exec("CREATE DATABASE \"$dbname\" WITH TEMPLATE = template0 encoding = 'UTF8' lc_collate='Spanish_Spain.1252' lc_ctype='Spanish_Spain.1252';");
                if($test === false)
                    throw new \Exception($db->errorInfo()[2]);
                $this->info(sprintf('Successfully created %s database', $dbname));
    }
    catch (\Exception $exception) {
                $this->error(sprintf('Failed to create %s database: %s', $dbname, $exception->getMessage()));
    }
}
}

它工作正常,但是我也想在该数据库中创建几个模式。我尝试在迁移文件中使用db :: unprepared,但是它不起作用,因为要进行迁移,它需要具有先前创建的架构。

EDIT1 :我尝试使用此迁移创建架构:

<?php

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

class CreateSchemaAdministracion extends Migration
{
/**
 * Run the migrations.
 *
 * @return void
 */
public function up()
{
    DB::unprepared('
        CREATE SCHEMA administracion
    ');
}

/**
 * Reverse the migrations.
 *
 * @return void
 */
public function down()
{
    DB::unprepared('DROP SCHEMA `administracion`');
}
}

但是我得到:无效的架构名称:7错误:未选择任何架构。

1 个答案:

答案 0 :(得分:0)

这将有助于:

        public function up()
    {
        DB::connection($this->getConnection())->unprepared("
        SET search_path to public;
        CREATE SCHEMA administracion;
        SET search_path to administracion;
    ");
    }

    public function down()
    {
        DB::connection($this->getConnection())->unprepared("
        DROP SCHEMA IF EXISTS administracion;
");
    }