每次用户第一次登录时,我都会保存一些关于他们的信息。如果用户已经有条目并且条目已过时,则应更新它。我的问题是试图将用户与数据库进行比较,看看他们是否已经有一个条目。我成功地记录了用户的详细信息,但每次用户登录时它都会这样做。您可以在下面找到我的代码。
$conn = new mysqli('host', 'user', 'password', 'db');
if ($conn->connect_error) {
die("Error; Contact Support!");
}
$stmt = $conn->prepare("select * from users where steamid=? ");
$stmt->bind_param('s', $steamprofile['steamid']);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($res);
$duplicate = mysqli_num_rows($res);
if ($duplicate == 0) {
$stmt1 = $conn->prepare("REPLACE INTO `users` SET `steamid` = ?,`realname` = ?,`username` = ?");
$stmt1->bind_param('sss', $steamprofile['steamid'], $steamprofile['realname'], $steamprofile['personaname']);
$stmt1->execute();
$stmt1->store_result();
$stmt1->bind_result($res1);
if ($res1 === TRUE) {
echo "";
}
$stmt1->close();
$stmt->close();
mysqli_close($conn);
}
最佳答案
您可以通过简单地使用 INSERT .. ON DUPLICATE KEY UPDATE 来避免所有这些复杂的逻辑:
$conn = new mysqli('host', 'user', 'password', 'db');
if ($conn->connect_error) {
die("Error; Contact Support!");
}
$query = "INSERT INTO `users` SET `steamid` = ?,`realname` = ?,`username` = ? ON DUPLICATE KEY UPDATE realname=? , username=?"
$stmt1 = $conn->prepare($query);
$stmt1->bind_param('sssss', $steamprofile['steamid'],
$steamprofile['realname'], $steamprofile['personaname'],
$steamprofile['realname'], $steamprofile['personaname']);
$stmt1->execute();
$stmt1->store_result();
$stmt1->bind_result($res1);
if ($res1 === TRUE) {
echo "";
}
$stmt1->close();
mysqli_close($conn);
您完全无需先运行查询来查明记录是否已存在。
另请注意,查询也可以写成
$query = "INSERT INTO `users` SET `steamid` = ?,`realname` = ?,`username` = ? ON DUPLICATE KEY UPDATE realname=VALUES(realname) , username=VALUES(username)"
在这种情况下,您不需要将用户名和真实姓名参数绑定(bind)两次。
请注意,这也优于 REPLACE INTO,因为替换会导致整行被替换,而不是仅更新一列。因此 REPLACE INTO 可能导致必须重新计算索引。
关于php - 将查询结果与当前用户进行比较,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38539369/