【发布时间】:2013-03-01 07:14:13
【问题描述】:
我正在尝试做的是INSERT 我的数据库中的订阅者,但IF EXISTS 它应该UPDATE 行,ELSE INSERT INTO 一个新行。
当然我首先连接到数据库,然后GET$name、$email 和$birthday 来自 url 字符串。
$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'];
这可行,但只是添加新行;
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);
但是它们都不起作用,我做错了什么?
非常感谢任何帮助!
【问题讨论】:
-
你的表有唯一约束吗?