Question

这不是一个重复的问题。上一个问题与PDO无关。我在mysql中有两个表：

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-（经理和工人都是员工）为了更新工人的保存字段（比如+ 10美元），现场工资需要更新 - $ 10

2-输入变量来自PHP表单name，因此逻辑流程为：

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

所以sql语句可以单独写成：

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;

是否可以在一个（PDO格式）中执行这3个语句。上面的msql PDO格式（用PHP）：

$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);

如何从另一个表更新mysql pdo中的字段

1 个答案: