laravel - 数据库播种器和 Laravel 的问题
问题描述
我有 3 个表,架构如下。雇员,
EmployeeDB.php
$factory->define(Employee::class, function (Faker $faker) {
return [
'emp_id' => $faker->randomDigit,
'name' => $faker->name,
'dept_id' => $faker->numberBetween($min = 1, $max = 15),
'salary_id' => $faker->randomDigit(),
'gender' => $faker->randomElement(['M', 'F', 'T']),
/*if I remove these next 2 statements, I receive an error for
SQLSTATE[42S22]: Column not found: 1054 Unknown column
'employee_id' in 'field list' (SQL: insert into `employees` (`emp_id`, `name`, `dept_id`, `salary_id`, `gender`, `date_of_joining`,
`date_of_birth`, `employee_id`, `updated_at`, `created_at`) values (2, Jamaal Beer, 3, 9, T, 2018-05-05 18:59:40, 2005-07-05 13:17:23, ?,
2019-12-11 11:15:42, 2019-12-11 11:15:42))
*/
'employee_id' => $faker->randomDigit,
//Same for this one as well
'department_id' => $faker->randomDigit,
'date_of_joining' => $faker->dateTimeBetween($startDate = '-5 years', $endDate = 'now', $timezone = null),
'date_of_birth' => $faker->dateTimeBetween($startDate = '-20 years', $endDate = 'now', $timezone = null),
员工迁移表:
Schema::create('employees', function (Blueprint $table) {
$table->bigIncrements('emp_id');
$table->integer('dept_id')->unsigned();
$table->foreign('dept_id')->references('id')->on('departments');
$table->integer('salary_id')->unsigned();
$table->foreign('salary_id')->references('id')->on('salaries');
$table->string('name');
$table->string('gender');
$table->timestamp('date_of_joining');
$table->dateTime('date_of_birth');
/*SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'employee_id' cannot be null
(SQL: insert into `employees` (`emp_id`, `name`, `dept_id`, `salary_id`, `gender`, `employee_id`, `department_id`,
`date_of_joining`, `date_of_birth`, `updated_at`, `created_at`)
values (6, Martina Wuckert, 7, 3, F, ?, 3, 2018-09-14 05:59:15, 20
*/
$table->string('employee_id')->nullable();
$table->string('department_id')->nullable();
$table->timestamps();
//added to prevent errors
部门 部门DB.php
factory->define(Department::class, function (Faker $faker) {
return [
//Yes, this is a hack. Use Multidimensional Arrays the next time.
'id' => $faker->numberBetween($min = 1, $max = 15),
'dept_name' => $faker->randomElement(['Production', 'Purchase & Quality', 'Operations', 'Sales', 'Customer Serice', 'Business Development', 'Maketing', 'Tech Support', 'Finance', 'Human Resources', 'Research & Development', 'IT', 'Legal']),
];
部门迁移
Schema::create('departments', function (Blueprint $table) {
$table->increments('id');
$table->string('dept_name');
$table->timestamps();
});
薪水
工资数据库.php
$factory->define(Salary::class, function (Faker $faker) {
return [
'id' => $faker->randomDigit(),
'monthlySalary' => $faker->randomNumber($nbDigits = 3),
//Yes this is a hack. Use MultiDimensional Arrays the next time.
];
});
工资移民
$factory->define(Salary::class, function (Faker $faker) {
return [
'id' => $faker->randomDigit(),
'monthlySalary' => $faker->randomNumber($nbDigits = 3),
//Yes this is a hack. Use MultiDimensional Arrays the next time.
];
});
App\Department 类 Department 扩展 Model
{
//
public function employee()
{
return $this->hasMany(Employee::class);
}
public function salary()
{
return $this->hasMany(Salary::class);
}
}
应用\员工
class Employee extends Model
{ //belongsto block
public function department()
{
return $this->hasOne(Department::class);
}
public function Salary()
{
return $this->hasOne(Salary::class);
}
}
应用\工资
class Salary extends Model
{
//
public function employee()
{
return $this->belongsTo(Employee::class);
}
}
DatabaseSeeder.php
factory(App\Employee::class, 25)->create()->each(function ($employee) {
$department = factory(App\Department::class)->make();
$employee->department()->save($department);
$salary = factory(App\Salary::class)->make();
$employee->salary()->save($salary);
});
当我运行以下命令时,这是我得到 php artisan db:seed 的错误
Integrity constraint violation: 1452 Cannot add or update a child row: a foreign key constraint fails (`volga_2`.`employees`, CONSTRAINT `employees_dept_id_foreign` FOREIGN KEY (`dept_id`) REFERENCES `departments` (`id`))")
我要做的是:为每份员工记录分配一个部门和一份薪水。
这是为了以后:部门不能在自己的表中重复。IE。一旦创建了带有名称的部门 id,就不能再次重新创建它。
我知道我错了,否则不会有错误。当我为部门和工资运行 dbSeeder 时,它运行良好。但是,只要有重复的部门条目,它就会引发错误。但是存储在数据库中的数据很好。
我想要一个解决方案或一些建议,因为我还在学习 Laravel。
谢谢!
编辑 当我执行 php artisan migrate
迁移顺序
Migration table created successfully.
Migrating: 2014_10_12_000000_create_users_table
Migrated: 2014_10_12_000000_create_users_table (0.44 seconds)
Migrating: 2014_10_12_100000_create_password_resets_table
Migrated: 2014_10_12_100000_create_password_resets_table (1.18 seconds)
Migrating: 2019_12_09_105432_create_departments_table
Migrated: 2019_12_09_105432_create_departments_table (0.29 seconds)
Migrating: 2019_12_09_105743_create_salaries_table
Migrated: 2019_12_09_105743_create_salaries_table (0.21 seconds)
Migrating: 2019_12_10_104739_create_employees_table
Migrated: 2019_12_10_104739_create_employees_table (2.04 seconds)
解决方案
只需更改迁移的顺序。我猜你正试图在创建表之前建立关系。进行父表的第一次迁移
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