[英]PHP: check row exist using mysqli query
我有这两个功能来创建用户。第一个功能用于插入/创建用户和其他功能用于检查用户是否存在于MySql
数据库中。 但是当我发送数据时,我看到die()的空白页; (如果用户已存在或用户不存在)如何解决此错误? 以及如何在红色框中打印错误列表,即:
- this user not activate
- email is empty
- ....
我的课:
public function createUser($email, $password, $salt, $verification) {
if($this->checkUserFound($email) != true ){die();}
$query = "INSERT INTO tbUsers (email, password, user_salt, is_verified, is_active, is_admin, verification_code) "
. "VALUES (?, ?, ?, ?, ?, ?, ?)";
$stmt = $this->_db->prepare($query);
$ver = 0;
$act = 1;
$adm = 0;
$stmt->bind_param("sssiiis", $email, $password, $salt, $ver, $act, $adm, $verification);
if ($stmt->execute()) {
return true;
}
return false;
}
public function checkUserFound($email) {
$query = "SELECT email FROM tbUsers where email = ?";
$stmt = $this->_db->prepare($query);
$stmt->bind_param("s", $email);
if ($stmt->execute()) {
$stmt->bind_result($email);
//fetch first row of results
$stmt->fetch();
return false;
}
return true;
}
这总是返回FALSE
(假设查询成功执行):
if ($stmt->execute()) {
$stmt->bind_result($email);
//fetch first row of results
$stmt->fetch();
return false;
}
您想要将其更改为:
if ($stmt->execute()) {
$stmt->store_result();
return 0 == $stmt->num_rows ? FALSE : TRUE;
}
然后,如果找到该用户,它将返回TRUE
; 如果不是,则返回FALSE
。
另外,我认为(从语义上)具有以下含义更有意义:
if($this->checkUserFound($email)){die();}
...,因为您希望它在找到用户后返回TRUE
(并结束脚本是因为您不想插入重复项)。/
if ($stmt->execute()) {
$stmt->bind_result($email);
//fetch first row of results
$stmt->fetch();
return true;
}
return false;
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