如何从另一个表中更新mysql pdo中的字段

Question

这不是重复的问题。前面的问题与PDO无关。我在mysql的两个表：

USERS 
------------------------------------- 
employeeid | name | saving | salary 
------------------------------------- 
12   | Bob | 100  | 1000 
23   | Joe | 50  | 800 

USERS table 
employeeid 
name 
saving 
salary 

和：

EMPLOYEE 
----------------------------------- 
id | managerid | workerid 
----------------------------------- 
1 | 12  | 23 

EMPLOYEE table 
id 
managerid FOREIGN KEY 
workerid FOREIGN KEY 

1-（既经理和工人都是员工），以更新保存领域的workerid（以+ 10 $说）字段薪水需要由被更新 - 10 $

2-输入变量来自PHP形式name，所以逻辑流程是：

name > find employeeid (id) from USERS > find managerid (id2) from EMPLOYEE > find employeeid (id3) from USERS > update saving and salary 

所以单独的SQL语句可以写成：

id = SELECT employeeid FROM USERS WHERE name = $name; //find id of employee in USERS 
id2 = SELECT managerid FROM EMPLOYEE WHERE workerid = id; //find id of worker in EMPLOYEE 
UPDATE USERS SET saving = saving + 10, salary = salary -10 WHERE employeeid = id2; 

是否有可能在一个做这3个报表（以PDO格式）。以上（与PHP） mSQL的PDO格式：

$sql = "SELECT employeeid FROM USERS WHERE name=:namepara"; 
$sttm = prepare($sql); 
$sttm->execute(array(":namepara"=>$name)); 
$row=$sttm->fetch(PDO::FETCH_ASSOC); 

$sql2 = "SELECT managerid FROM EMPOYEE WHERE workerid=:idpara"; 
$sttm2 = prepare($sql2); 
$sttm2->execute(array(":idpara"=>$row['employeeid'])); 
$row2=$sttm2->fetch(PDO::FETCH_ASSOC); 

$sql3 = "UPDATE USERS SET saving = saving + 10, salary = salary - 10 WHERE 
    employeeid=:id2para"; 
$sttm3 = prepare($sql3); 
$sttm3->execute(array(":id2para"=>$row2['managerid'])); 
$row3=$sttm3->fetch(PDO::FETCH_ASSOC); 

任何帮助，将不胜感激！

Answer 1

$sql = "UPDATE USERS JOIN EMPLOYEE ON USERS.employeeid = EMPLOYEE.managerid 
     SET Saving = saving +10, salary = salary - 10 
     WHERE USERS.name = :namepara"; 

$sttm = prepare($sql); 
$sttm->execute(array(":namepara"=>$name)); 
$row=$sttm->fetch(PDO::FETCH_ASSOC);