[英]bind_param error
This is my Error:这是我的错误:
Warning: mysqli_stmt::bind_param() [mysqli-stmt.bind-param]: Number of variables doesn't match number of parameters in prepared statement in /Applications/XAMPP/xamppfiles/htdocs/Jil/benutzer_eintragen.php on line 19
警告:mysqli_stmt::bind_param() [mysqli-stmt.bind-param]:变量数与第 19 行 /Applications/XAMPP/xamppfiles/htdocs/Jil/benutzer_eintragen.php 中准备好的语句中的参数数不匹配
and this is my Code这是我的代码
$sql = "INSERT INTO benutzer SET vorname='?', nachname='?', username='?', email='?', passwort='?';";
$stmt = $db->prepare($sql);
$stmt->bind_param("sssss", $vorname, $nachname, $username, $email, $passwort);
$stmt->execute();
I think you need to eliminate the quote marks in the statement:我认为您需要删除语句中的引号:
$sql = "INSERT INTO benutzer SET vorname=?, nachname=?, username=?, email=?, passwort=?;";
You don't need the quotes around the ?
您不需要 ? 周围的引号
?
in the SQL statement.在 SQL 语句中。
Also, your SQL statement is incorrect.
此外,您的 SQL 语句不正确。
SET
is only used with
UPDATE
,
INSERT
uses
VALUES
.
SET
仅用于
UPDATE
,
INSERT
使用
VALUES
。
$sql = "INSERT INTO benutzer(vorname,nachname,username,email,passwort) VALUES (?,?,?,?,?)";
$sql = "INSERT INTO benutzer SET vorname=?, nachname=?, username=?, email=?, passwort=?";
$stmt = $db->prepare($sql);
$stmt->bind_param("sssss", $vorname, $nachname, $username, $email, $passwort);
$stmt->execute();
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