SQL -如果存在UPDATE ELSE INSERT
SQL -如果存在UPDATE ELSE INSERT
提问于 2013-03-13 19:21:05
我正在尝试做的是在我的数据库中的INSERT订阅者，但它应该UPDATE这一行，ELSE INSERT INTO一个新的行。
当然，我首先连接到数据库，并从url字符串中GET出$name、$email和$birthday。
$con=mysqli_connect("localhost","---","---","---");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name=$_GET['name'];
$email=$_GET['email'];
$birthday=$_GET['birthday'];这是可行的，但只是添加了新的行；
mysqli_query($con,"INSERT INTO subs (subs_name, subs_email, subs_birthday)
VALUES ('$name', '$email', '$birthday')");
mysqli_close($con);这是我尝试过的；
mysqli_query($con,"INSERT INTO subs (subs_name, subs_email, subs_birthday)
VALUES '$name', '$email', '$birthday'
ON DUPLICATE KEY UPDATE subs_name = VALUES($name), subs_birthday = VALUES($birthday)");
mysqli_close($con);和
mysqli_query($con,"IF EXISTS (SELECT * FROM subs WHERE subs_email='$email')
UPDATE subs SET subs_name='$name', subs_birthday='$birthday' WHERE subs_email='$email'
ELSE
INSERT INTO subs (subs_name, subs_email, subs_birthday) VALUES ('$name', '$email', '$birthday')");
mysqli_close($con);和
mysqli_query($con,"IF NOT EXISTS(SELECT * FROM subs WHERE subs_email='$email')
Begin
INSERT INTO subs (subs_name, subs_email, subs_birthday)
VALUES ('$name', '$email', '$birthday')
End");
mysqli_close($con);但是它们都不工作，我做错了什么？
任何帮助都是非常感谢的！
回答 1
Stack Overflow用户
发布于 2020-10-16 12:34:04
试试这个：
INSERT INTO `center_course_fee` (`fk_course_id`,`fk_center_code`,`course_fee`) VALUES ('69', '4920153', '6000') ON DUPLICATE KEY UPDATE `course_fee` = '6000';页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/15383852
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