Question

当我运行此代码时，我得到无法添加或更新子行：外键约束失败。我需要帮助来解决这个问题。 include_once（＆＃39;的config.php＆＃39）;

- 第一张表 -

CREATE TABLE welfare_funeral(
welfare_id INT AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255),
Date Date,
Venue VARCHAR(255)
)ENGINE= INNODB;

- 第二张表 -

CREATE TABLE welfare_funeral_payees(
payee_id INT AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255),
Amount VARCHAR(255),
welfare_id INT,
INDEX wel_id(welfare_id),
FOREIGN KEY(welfare_id) REFERENCES welfare_funeral(welfare_id)
)ENGINE=INNODB;

- PHP代码 -

if(isset($_POST['record'])){
    $payeeName = $_POST['payeeName'];
    $amount = $_POST['amount'];
    $welfare_id="SELECT MAX('welfare_id') FROM 'welfare_funeral'";

    $myd = $pdo->prepare('INSERT INTO welfare_funeral_payes(Name, Amount, welfare_id)VALUES(:upayeeName, :uamount, :uwelfare_id)');
    $myd->bindParam(':upayeeName', $payeeName);
    $myd->bindParam(':uamount', $amount);
    $myd->bindParam(':uwelfare_id', $welfare_id);

    if($myd->execute()){
        ?>
        <script>
            alert("New Welfare created");
        </script>
        <?php
    }else{
        ?>
        <script>
            alert("Couln't not create Welfare");
        </script>
        <?php
    }
}

Answer 1

您不能将subselect作为参数放在预准备的Statement中。

您必须首先查询id，然后您可以将这些设置为参数，或者将子选择直接写入语句：

    $myd = $pdo->prepare('INSERT INTO welfare_funeral_payes(Name, Amount, welfare_id) SELECT :upayeeName,uamount, MAX(welfare_id) FROM welfare_funeral);
    $myd->bindParam(':upayeeName', $payeeName);
    $myd->bindParam(':uamount', $amount);
    $myd->bindParam(':uwelfare_id', $welfare_id);

也可以使用反引号arround列和表名而不是单引号

外键约束失败

1 个答案: