无法使用 Laravel eloquent ORM 插入 PostgreSQL
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Unable to insert into PostgreSQL using laravel eloquent ORM
提问by Mo Fetch
I did my research a lot before posting this.
在发布这篇文章之前，我做了很多研究。
I am getting the following error when trying to insert a record into PostgreSQL using Laravel eloquent ORM.
尝试使用 Laravel eloquent ORM 将记录插入 PostgreSQL 时出现以下错误。
This is the error:
这是错误：
Illuminate \ Database \ QueryException
SQLSTATE[42703]: Undefined column:
7 ERROR: column "id" does not exist LINE 1: ...ed_at", "created_at")
values (, , , )
returning "id" ^ (SQL: insert into "fb_post_like_counts" ("post_id", "count", "updated_at", "created_at") values (293843523498_10152007721598499, 0, 2014-03-12 16:56:24, 2014-03-12 16:56:24) returning "id")
This is the database table create schema:
这是数据库表创建模式：
Schema::create('fb_post_shares_counts', function($table)
{
//
$table->string('post_id');
$table->string('count')->default(0);
// created_at updated_at deleted_at
$table->timestamps();
$table->softDeletes();
// set composite keys
$table->primary(array('post_id', 'updated_at'));
});
and this is the code i am trying to execute:
这是我要执行的代码：
// SAVE POST SHARES
$post_share_count = new FbPostShareCount;
$post_share_count->post_id = $response['id'];
$post_share_count->count = $fql_post['share_count'];
$post_share_count->save();
and I created a model class FbPostShareCount extends Eloquent.
我创建了一个模型类 FbPostShareCount 扩展了 Eloquent。
I know that laravel does not support complex keys but this problem did not occur when I was using MySQL
我知道laravel不支持复杂的键但是我在使用MySQL时没有出现这个问题
回答by Antonio Carlos Ribeiro
Set the primary key in your FbPostShareCountmodel as
将FbPostShareCount模型中的主键设置为
class FbPostShareCount extends Eloquent {
protected $primaryKey = 'post_id';
...
}
回答by arcsum
have you tried using the correct data type for post id?
您是否尝试过为帖子 ID 使用正确的数据类型？
$table->integer('post_id');
回答by Chando
Laravel eloquent ORM have auto increment, So set up primaryKey = null and increment = false.
Laravel eloquent ORM 有自动增量，所以设置primaryKey = null 和increment = false。
protected $primaryKey = null;
public $incrementing = false;
