表一:服务, 表二:serviceImg
服务表具有一列作为serviceImg。服务表的此serviceImg列等于serviceImg表的serviceImgId。因此,我已经完成了这些表之间的内部联接。
摘要: 简而言之,我想从两个表中获取数据,插入两个表中,一次更新两个表。
我该怎么做?
这是我的模特
public function update_services($serviceId,$data)
{
$this->db->select('
serviceImg.serviceImgId,
serviceImg.serviceImgName,
service.serviceId,
service.serviceType,
service.serviceCat,
service.serviceNameEn,
service.serviceNameAr,
service.serviceDescEn,
service.serviceDescAr,
service.servicePrice,
service.servicePack,
service.serviceCreateUser,
service.serviceCreateDate,
service.serviceEditUser,
service.serviceEditDate,
service.serviceImg,
service.serviceStatus,
serviceCat.serviceCatId,
serviceCat.serviceCatNameEn,
serviceCat.serviceCatNameAr,
serviceCat.serviceCatType,
serviceCat.serviceCatCreateUser,
serviceCat.serviceCatCreateDate,
serviceCat.serviceCatEditUser,
serviceCat.serviceCatEditDate,
serviceCat.serviceCatActive
');
$this->db->from('serviceImg');
$this->db->set('serviceImg.serviceImgId = service.serviceImg');
$this->db->join('service', 'serviceImg.serviceImgId = service.serviceImg', 'LEFT');
$this->db->join('serviceCat', 'service.serviceCat = serviceCat.serviceCatId', 'LEFT');
$this->db->where('serviceId', $serviceId);
$this->db->where('service.serviceImg = serviceImg.serviceImgId');
$this->db->update('service', $data);
}
我在等待你的答案。
答案 0 :(得分:0)
您可以使用codeigniter的TRANSACTION
$this->db->trans_start();
$this->db->query('AN SQL QUERY...');
$this->db->query('ANOTHER QUERY...');
$this->db->query('AND YET ANOTHER QUERY...');
$this->db->trans_complete();
这是link
或者您可以使用ajax插入或更新多个表。