php - 错误插入 ON duplicatekey UPDATE 函数

Question

我在数据库中有一个表“fuser”，其中提到了各自的字段。我有带有字段输入区域的 .php 表单。我尝试了 INPUT INTO & DUPLICATEKEY UPDATE 函数，但它似乎出现了一些错误，例如 ERRORINSERTINTO 你的 SQL 语法有错误；查看与您的 MariaDB 服务器版本对应的手册，了解在附近使用的正确语法

我的代码如下:

<?php include('profile.php'); ?>

<?php

$servername  = "localhost";
$dbusername = "root";
$dbpassword = "*******";
$dbname = "the_database";

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$aboutme = $_POST['aboutme'];
$subject1 = $_POST['subject1'];
$subject2 = $_POST['subject2'];
$subject3 = $_POST['subject3'];
$country = $_POST['country'];
$birthday = $_POST['birthday'];
$occupation = $_POST['occupation'];
$mobile = $_POST['mobile'];
$websiteurl = $_POST['websiteurl'];



$conn = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}


$sql = "INSERT INTO fuser (firstname, lastname, aboutme, subject1, subject2, subject3, country, birthday, occupation, mobile, websiteurl)
VALUES ('$firstname', '$lastname', '$aboutme', '$subject1', '$subject2', '$subject3', '$country', '$birthday', '$occupation', '$mobile', '$websiteurl')
ON DUPLICATE KEY UPDATE
  firstname     = VALUES('$firstname'),
  lastname      = VALUES('$lastname'),
  aboutme       = VALUES('$aboutme'),
  subject1      = VALUES('$subject1'),
  subject2      = VALUES('$subject2'),
  subject3      = VALUES('$subject3'),
  country       = VALUES('$country'),
  birthday      = VALUES('$birthday'),
  occupation    = VALUES('$occupation'),
  mobile        = VALUES('$mobile'),
  websiteurl    = VALUES('$websiteurl')";

if ($conn->query($sql) === TRUE) {
echo '<script language="javascript">';
echo 'alert("Your details have been updated succesfully..")';
echo '</script>';
echo '<a href="profile.php"></a>';
}
else {
    echo "ERROR" . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

感谢任何建议..

Answer 1

我的错，@furrie 是对的..代码应该是

ON DUPLICATE KEY UPDATE
  firstname     = VALUES(firstname),
  lastname      = VALUES(lastname),
  aboutme       = VALUES(aboutme),
  subject1      = VALUES(subject1),
  subject2      = VALUES(subject2),
  subject3      = VALUES(subject3),
  country       = VALUES(country),
  birthday      = VALUES(birthday),
  occupation    = VALUES(occupation),
  mobile        = VALUES(mobile),
  websiteurl    = VALUES(websiteurl)";

现在它像 gem 一样工作......

没关系.. 我实际上需要使用 UPDATE-SET 条件而不是这个.. 无论如何感谢大家的时间.. 通过这个我刚刚学到了一些东西，我们不能将 WHERE 条件与 INSERT INTO 一起使用...