架构构建器laravel迁移在两列上唯一

Question

如何在两列上设置唯一约束？

class MyModel extends Migration {
  public function up()
  {
    Schema::create('storage_trackers', function(Blueprint $table) {
      $table->increments('id');
      $table->string('mytext');
      $table->unsignedInteger('user_id');
      $table->engine = 'InnoDB';
      $table->unique('mytext', 'user_id');
    });
  }
}

MyMode::create(array('mytext' => 'test', 'user_id' => 1);
// this fails??
MyMode::create(array('mytext' => 'test', 'user_id' => 2);

Answer 1

第二个参数是手动设置唯一索引的名称.使用数组作为第一个参数,可以跨多个列创建唯一键.

$table->unique(array('mytext', 'user_id'));

或者(有点整洁)

$table->unique(['mytext', 'user_id']);

Answer 2

如果您有一个包含一列的默认唯一索引，并且您将使用两列更改它，或者创建一个包含两列的新索引，则此脚本将为您执行此操作：

public function up()
{
    Schema::table('user_plans', function (Blueprint $table) {
        $table->unique(["email", "plan_id"], 'user_plan_unique');
    });
}

public function down()
{
    Schema::table('user_plans', function (Blueprint $table) {
      $table->dropUnique('user_plan_unique');
    });
}

Answer 3

只需你可以使用

$table->primary(['first', 'second']);

参考:http://laravel.com/docs/master/migrations#creating-indexes

举个例子:

    Schema::create('posts_tags', function (Blueprint $table) {

        $table->integer('post_id')->unsigned();
        $table->integer('tag_id')->unsigned();

        $table->foreign('post_id')->references('id')->on('posts');
        $table->foreign('tag_id')->references('id')->on('tags');

        $table->timestamps();
        $table->softDeletes();

        $table->primary(['post_id', 'tag_id']);
    });