您好，所有開發人員，我有下一個查詢，根據我所在的國家/地區，我需要從所選字段中刪除一個我不需要的元素，因為在某些國家/地區的數據庫中，該特定列（字段）不存在。在其他國家/地區，查詢必須保持初始狀態。因此，查詢將遵循以下結構： $options['joins'] = array( array( 'table' => 'td_addresses', 'alias' => 'Address', 'type' => 'LEFT', 'conditions' => array( 'Doctor.id = Address.doctor_id' ) ), array( 'table' => 'td_doctors_products', 'alias' => 'DoctorsProducts', 'type' => 'LEFT', 'conditions' => array( 'Doctor.id = DoctorsProducts.id_doctor' ) ), array( 'table' => 'td_products', 'alias' => 'Products', 'type' => 'LEFT', 'conditions' => array( 'DoctorsProducts.id_product = Products.id' ) ), array( 'table' => 'td_addresses_agenda', 'alias' => 'AddressesAgenda', 'type' => 'LEFT', 'conditions' => array( 'AddressesAgenda.address_id = Address.id' ) ));$options['conditions'] = array( 'Doctor.email !=' => '', 'Doctor.accepted_social_politics >=' => 0);$options['fields'] = array( 'Doctor.id', 'Doctor.email', 'Doctor.name', 'Doctor.surname', 'Doctor.created', 'Doctor.profileimg', 'Doctor.list_experiencia_professional', 'Doctor.logros_academicos', 'Doctor.premios_reconocimientos', 'Doctor.status', 'Doctor.type_doctor', 'Doctor.sexo', 'Doctor.accepted_politics_wallet', 'Doctor.accepted_social_politics', 'DoctorsProducts.id_product', 'Address.phone', 'Address.info_consulta', 'DoctorsProducts.status', 'AddressesAgenda.address_id');echo json_encode($options['fields']["Doctor.accepted_politics_wallet"]);$latam = ['mx', 'co'];if(in_array(PAIS, $latam)){ // Remove the field of Doctor.accepted_politics_wallet from $options['fields']}$options['order'] = array('Doctor.id ASC');$options['group'] = array('Doctor.email');$doctors_csv = $this->Doctor->find('all', $options);應用 array_splice 是否可行，對嗎？提前致謝
2 回答
qq_花開花謝_0
TA貢獻1835條經驗 獲得超7個贊
實際上有很多方法可以使用現有的 php 數組函數，例如，array_filter()等。array_diff()array_search()array_search()
if (($key = array_search('Doctor.accepted_politics_wallet', $options['fields'])) !== false) {
unset($options['fields'][$key]);
}
print_r($options);
工作演示： https ://3v4l.org/nkQZt
- 2 回答
- 0 關注
- 128 瀏覽
添加回答
舉報
0/150
提交
取消