Question

我正在用PHP设置一个简单的CRUD，我在phpmyadmin中创建了两个表，一个表用于个人信息（姓名，姓氏，电子邮件，电话），另一个表用于其他电话号码。我在设置外键来连接两个表时遇到问题。附加电话号码是在jQuery的帮助下从用户动态添加的（可以根据用户是否拥有两个电话号码来调用该附加电话号码）。有人可以指出我们在哪里犯错。我对PHP的了解有限，我从phpmyadmin的GUI创建表和数据库，而没有使用代码。

用于在数据库中输入添加的表单的代码。

// Checks if the submit button has been pressed, setst the variable values of the name values from the form
if (isset($_POST['submit'])){
  $first_name = $_POST['first_name'];
  $last_name = $_POST['last_name'];
  $email = $_POST['email'];
  $phone_number = $_POST['phone_number'];
$mysqli->query("INSERT INTO personal (first_name, last_name, email, phone_number) VALUES('$first_name', '$last_name', '$email', '$phone_number')")
      OR die($mysqli->error);
}

// Checks if the Add more numbers button has been pressed
 if (isset($_POST['phone_number2'])) {
   $phone_number2 = $_POST['phone_number2'];
// foreach statement to loop trought the array of inserted additional phone numbers
  foreach ($phone_number2 as $key => $value) {
      $mysqli->query("INSERT INTO phone (phone_number2) VALUES('$value')")
               OR die($mysqli->error);
}
}

直到我输入外键为止，它一直有效，输入后我得到了错误。无法添加或更新子行：外键约束失败（exercise。phone，CONSTRAINT phone_ibfk_1外部键（user_id）参考personal（{{ 1}}））

Answer 1

使用$mysqli->insert_id获取在上一个查询中添加的用户ID，并将其用作phone表中的外键。

if (isset($_POST['submit'])){
    $first_name = $_POST['first_name'];
    $last_name = $_POST['last_name'];
    $email = $_POST['email'];
    $phone_number = $_POST['phone_number'];
    $mysqli->query("INSERT INTO personal (first_name, last_name, email, phone_number) VALUES('$first_name', '$last_name', '$email', '$phone_number')")
        OR die($mysqli->error);
    $userid = $mysqli->insert_id;
// Checks if the Add more numbers button has been pressed
    if (isset($_POST['phone_number2'])) {
        $phone_number2 = $_POST['phone_number2'];
// foreach statement to loop trought the array of inserted additional phone numbers
        foreach ($phone_number2 as $key => $value) {
            $mysqli->query("INSERT INTO phone (userid, phone_number2) VALUES($userid, '$value')")
                OR die($mysqli->error);
        }
    }
}

如何修复MySQL中的“外键约束失败”

1 个答案: