我正在尝试运行以下迁移：
public function up()
{
Schema::create('lifestyle_questions', function(Blueprint $table)
{
$table->increments('id');
$table->string('question');
$table->timestamps();
});
Schema::create('lifestyle_question_answers', function(Blueprint $table)
{
$table->increments('id');
$table->integer('lifestyle_question_id')->unsigned();
$table->foreign('lifestyle_question_id')->references('id')->on('lifestyle_questions');
$table->string('answer');
$table->timestamps();
});
Schema::create('user_lifestyle_question_answers', function(Blueprint $table)
{
$table->integer('user_id')->unsigned();
$table->foreign('user_id')->references('id')->on('users');
$table->integer('lifestyle_question_answer_id')->unsigned();
$table->foreign('lifestyle_question_answer_id')->references('id')->on('lifestyle_question_answers');
});
}
但我收到以下错误：
[Illuminate\Database\QueryException]
SQLSTATE[42000]: Syntax error or access violation: 1059 Identifier name 'user_lifestyle_question_answers_lifestyle_question_answer_id_foreign' is too long (SQL: alter table `user_lifestyle_question_answers` add constraint user_lifestyle_question_answers_lifestyle_question_answer_id_foreign foreign key (`lifestyle_question_answer_id`) references `lifestyle_question_answers` (`id`))
[PDOException]
SQLSTATE[42000]: Syntax error or access violation: 1059 Identifier name 'user_lifestyle_question_answers_lifestyle_question_answer_id_foreign' is too long
C
Chuck Le Butt
您可以将自定义索引名称作为第二个参数传递给 foreign() 方法。或者只使用较短的表/列名。
所以你想做这样的事情：
public function up()
{
Schema::create('lifestyle_questions', function(Blueprint $table)
{
$table->increments('id');
$table->string('question');
$table->timestamps();
});
Schema::create('lifestyle_question_answers', function(Blueprint $table)
{
$table->increments('id');
$table->integer('lifestyle_question_id')->unsigned();
$table->foreign('lifestyle_question_id', 'lq_id_foreign')->references('id')->on('lifestyle_questions');
$table->string('answer');
$table->timestamps();
});
Schema::create('user_lifestyle_question_answers', function(Blueprint $table)
{
$table->integer('user_id')->unsigned();
$table->foreign('user_id')->references('id')->on('users');
$table->integer('lifestyle_question_answer_id')->unsigned();
$table->foreign('lifestyle_question_answer_id', 'lqa_id_foreign')->references('id')->on('lifestyle_question_answers');
});
}
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primary()方法：$table->primary(['id1', 'id'], 'custom_index');如何解决此问题？foreignId的人，您可以像这样使用它：$table->foreignId('really_long_foreign_key_id')->constrained()->index('short_id_foreign');