我有2个表组关系杂货店CRUD错误
CREATE TABLE `tbl_patient` (
id_patient INTEGER NOT NULL PRIMARY KEY AUTO_INCREMENT,
name VARCHAR(25) NOT NULL DEFAULT "not available",
att1 VARCHAR(5) NOT NULL DEFAULT "att1",
att2 VARCHAR(25) NOT NULL DEFAULT "att2",
att3 VARCHAR(25) NOT NULL DEFAULT "att3",
CONSTRAINT `uc_Info_patient` UNIQUE (`id_patient`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1;
和
CREATE TABLE `tbl_patient_medicine` (
id_patient_medicine INTEGER NOT NULL PRIMARY KEY AUTO_INCREMENT,
id_patient INTEGER NOT NULL,
name_medicine VARCHAR(50) NOT NULL DEFAULT "",
dosis VARCHAR(50) NOT NULL DEFAULT "",
start_date timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
treatment VARCHAR(50) NOT NULL DEFAULT "",
times_per_day VARCHAR(50) NOT NULL DEFAULT "",
CONSTRAINT fk_ID_Patient_Medicine FOREIGN KEY (id_patient) REFERENCES `tbl_patient`(id_patient)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1;
正如你可以看到表patient_medicine是tbla_medicines之间的中间表,表病人。
现在我想用杂货污物协商从tbl_patient_medicine所有数据like in this sqlfiddle
supossing我通过在URI中的ID(在本例将id_patient = 1) 我有
public function details_medication($my_id = 0)
{
try{
$crud = new grocery_CRUD();
$crud->where('id_patient',$my_id);
$crud->set_table('tbl_patient_medicine');
//HOW TO DO IT?
$crud->set_relation('id_patient', 'tbl_patient', 'id_patient');
$output = $crud->render();
$this->_output($output);
}catch(Exception $e){
show_error($e->getMessage().' --- '.$e->getTraceAsString());
}
}
SO我尝试了不同的方式,但得到这样的错误:
A Database Error Occurred
Error Number: 1052
Column 'id_patient' in where clause is ambiguous
SELECT `tbl_patient_medicine`.*, j7a675883.id_patient AS s7a675883
FROM (`tbl_patient_medicine`)
LEFT JOIN `tbl_patient` as j7a675883
ON `j7a675883`.`id_patient` = `tbl_patient_medicine`.`id_patient` WHERE `id_patient` = '1' LIMIT 25
行号:87
我不知道,我想这就是它使用不同的列名引起problem.Try的主键,试试这个,$ crud-> set_relation( 'id_patient',' tbl_patient','name');试试n temme O/p – Christopher 2013-03-19 09:27:17