我想创建一个将删除表的迁移。我这样创建了迁移:
Schema::table('devices', function (Blueprint $table) {
$table->increments('id');
$table->unsignedInteger('client_id')->nullable();
$table->foreign('client_id')->references('id')->on('clients')->onDelete('cascade');
});
现在我尝试像这样删除它:
Schema::table('devices', function (Blueprint $table) {
$table->dropForeign('devices_client_id_foreign');
$table->drop('devices');
});
但是出现以下错误:
In Connection.php line 664: SQLSTATE[42000]: Syntax error or access violation: 1091 Can't DROP 'devices_client_id_foreign'; check that column/key exists (SQL:
更改表
devices
删除外键devices_client_id_foreign
)In PDOStatement.php line 144: SQLSTATE[42000]: Syntax error or access violation: 1091 Can't DROP 'devices_client_id_foreign'; check that column/key exists In PDOStatement.php line 142: SQLSTATE[42000]: Syntax error or access violation: 1091 Can't DROP 'devices_client_id_foreign'; check that column/key exists
答案 0 :(得分:4)
您可以使用此答案。 https://stackoverflow.com/a/30177480/8513937
将列名作为数组传递给 dropForeign 。在内部,Laravel删除关联的外键。
$table->dropForeign(['client_id']);
答案 1 :(得分:3)
只需删除整个表格(documentation):
Schema::drop('devices');
答案 2 :(得分:1)
您只需要在删除表之前禁用外键检查,然后按如下所示再次启用它们:
DB::statement('SET FOREIGN_KEY_CHECKS=0;');
Schema::dropIfExists('devices');
DB::statement('SET FOREIGN_KEY_CHECKS=1;');
答案 3 :(得分:0)
尝试这种方式...
public function down()
{
Schema::dropIfExists('devices');
}
//Or this
public function down(){
Schema::table('devices', function (Blueprint $table) {
$table->dropForeign(['client_id']);
$table->dropColumn('client_id');
$table->drop('devices');
});
}