Estoy tratando de ejecutar la siguiente migración:
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');
});
}
Pero me sale el siguiente error:
[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
Pawel Bieszczad
Puede pasar un nombre de índice personalizado como segundo parámetro en el método Foreign(). O simplemente use nombres de tabla/columna más cortos.
Así que quieres hacer algo como:
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();
$table->foreign('lifestyle_question_answer_id', 'lqa_id_foreign')->references('id')->on('lifestyle_question_answers');
});
}
-
Solo para información: los identificadores de MySQL siempre deben tener menos de 64 caracteres según dev.mysql.com/doc/refman/5.5/en/identificadores.html
– Pᴇʜ
11 mayo 2015 a las 14:37
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Desafortunadamente, esta solución no funciona con el
primary()
método al generar claves compuestas:$table->primary(['id1', 'id'], 'custom_index');
¿Cómo resolver este problema?– Dong3000
8 de mayo de 2018 a las 13:04
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Para aquellos que tienen este problema al generar claves compuestas como @Dong3000, vean esto: stackoverflow.com/a/28626907/3368784
– Juana
16 mayo 2019 a las 18:40
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Para las personas que usan
foreignId
puedes usarlo así:$table->foreignId('really_long_foreign_key_id')->constrained()->index('short_id_foreign');
– Fredrik Jungstedt
12 de junio de 2020 a las 13:11