[英]Laravel create simple drop-down from relationship and concat columns
我想從2個具有單一關系的表的HTML代碼中創建簡單的下拉列表,在我的代碼中,我無法獲取關系數據以與其他表列連接。 例如,這是我的代碼:
$user_accounts = UserAccountNumber::with('currencyType')->select('*', DB::raw('CONCAT("CardNumber: ", card_number) AS account_info'))
->whereUserId(Auth::user()->id)
->pluck('account_info', 'id');
這段代碼可以正常工作,但是我想用UserAccountNumber
表來配置一些currencyType
表列,並且我不能使用tableName.columnName進入DB::raw()
例如:
$user_accounts = UserAccountNumber::with('currencyType')
->select(
'*',
DB::raw('CONCAT(" CardNumber: ", card_number, "CurrencyType: ", currencyType.title) AS account_info'))
->whereUserId(Auth::user()->id)
->pluck('account_info', 'id');
然后我得到這個錯誤:
SQLSTATE[42S22]: Column not found: 1054 Unknown column 'currencyType.currency_type' in 'field list' (SQL: select *, CONCAT("AccountNumber: ",account_number, " CardNumber: ", card_number, "CurrencyType: ", currencyType.currency_type) AS account_info from `user_account_numbers` where `user_id` = 17)
Model中的currencyType
方法:
public function currencyType()
{
return $this->belongsTo(CurrencyType::class, 'currency_type', 'id');
}
Laravel不使用JOIN進行關系,因此您必須為此編寫一個特定的查詢。 為此使用查詢生成器。
這樣的事情應該起作用:
$user_accounts = DB::table('user_account_numbers')
->join('currencyType', 'user_account_numbers.currency_id', '=', 'currencyType.id')
->select('*', DB::raw('CONCAT(" CardNumber: ", card_number, "CurrencyType: ", currencyType.title) AS account_info'))
->whereUserId(Auth::user()->id)
->get();
希望您能想到這個想法,並可以使其適應您的數據庫架構。
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