使用PHP PDO,有两个语句,想法是两个stmt1 并且stmt2是真的它应该提交否则它应该回滚,但正如我在这里看到的那样它没有得到回滚,如果stmt1为真,它将注释即使stmt2为假。
这是功能:
public function insert() {
// try { $stmt1 = $this->conn->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$this->conn->beginTransaction();
$stmt1 = $this->conn->prepare("INSERT into table1 (item,itemname,price)VALUES (:name, :itemname, :price)");
$stmt1->bindParam(':name' , $this->name);
$stmt1->bindParam(':itemname' , $this->itemname);
$stmt1->bindParam(':price' , $this->price);
$stmt1->execute();
$stmt2 = $this->conn->prepare("INSERT into table2 (item,itemname,price) VALUES (:name, :itemname, :price)");
$stmt2->bindParam(':name' , $this->name);
$stmt2->bindParam(':itemname' , $this->itemname);
$stmt2->bindParam(':price' , $this->price);
$stmt2->execute();
//} catch(PDOException $r){ echo $r->__toString();exit; }
if($stmt1 && $stmt2){
$this->conn->commit(); //This will save changes
} else {
$this->conn->rollBack(); //This will undo changes
}
}
}
这里我检查了这个函数,如果stmt1为true,它将运行并将数据插入表一,即使stmt2为false
问题:如何保持运行stmt1然后运行stmt2,如果stmt1为false,则不应运行stmt2,如果stmt2为false,则还应回滚stmt1。
提前感谢。
答案 0 :(得分:2)
试试这个:
public function insert() {
// try { $stmt1 = $this->conn->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$this->conn->beginTransaction();
$stmt1 = $this->conn->prepare("INSERT into table1 (item,itemname,price)VALUES (:name, :itemname, :price)");
$stmt1->bindParam(':name' , $this->name);
$stmt1->bindParam(':itemname' , $this->itemname);
$stmt1->bindParam(':price' , $this->price);
//$stmt1->execute();
$stmt2 = $this->conn->prepare("INSERT into table2 (item,itemname,price) VALUES (:name, :itemname, :price)");
$stmt2->bindParam(':name' , $this->name);
$stmt2->bindParam(':itemname' , $this->itemname);
$stmt2->bindParam(':price' , $this->price);
//$stmt2->execute();
//} catch(PDOException $r){ echo $r->__toString();exit; }
if($stmt1->execute() && $stmt2->execute()){
$this->conn->commit(); //This will save changes
} else {
$this->conn->rollBack(); //This will undo changes
}
}
}