我正在尝试获取2个表,它们都具有名为email
password
和isBusiness
的列,并在一个名为authentication
的表中自动更新这些值
认证迁移
Schema::create('authentication', function (Blueprint $table) {
$table->increments('auth_id');
$table->string('email');
$table->foreign('email')->references('email')->on('businesses');
$table->foreign('email')->references('email')->on('consumers');
$table->string('password');
$table->foreign('password')->references('password')->on('businesses');
$table->foreign('password')->references('password')->on('consumers');
$table->integer('isBusiness');
$table->foreign('isBusiness')->references('isBusiness')->on('businesses');
$table->foreign('isBusiness')->references('isBusiness')->on('consumers');
$table->timestamps();
});
错误
SQLSTATE[HY000]: General error: 1005 Can't create table `sprout_db`.`#sql-2
7a4_30` (errno: 150 "Foreign key constraint is incorrectly formed")
业务表
Schema::create('businesses', function (Blueprint $table) {
$table->increments('bus_id', 11);
$table->string('bus_name', 50);
$table->string('bus_address', 50);
$table->string('bus_city', 50);
$table->string('bus_prov', 50);
$table->string('bus_postal', 50);
$table->string('bus_phone', 50);
$table->string('email', 50);
$table->string('password', 100);
$table->integer('isBusiness')->default('1');
$table->timestamps();
$table->rememberToken();
$table->engine = 'InnoDB';
});
消费者表
Schema::create('consumers', function (Blueprint $table) {
$table->increments('con_id', 11);
$table->string('con_fname', 50);
$table->string('con_lname', 50);
$table->string('email', 50);
$table->string('password', 100);
$table->integer('isBusiness')->default('0');
$table->timestamps();
$table->rememberToken();
$table->engine = 'InnoDB';
});
这是正确的做法吗?通过为每列分配2个外键?我是否只有语法错误,或者这样做是不可能的?
最佳答案
只需将index()方法添加到您的密钥即可。
$table->integer('isBusiness')->index()->unsigned();
关于php - 使列引用来自单独表的2列Laravel,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/50769174/