When I run this code I give this error:
mysqli_error() expects exactly 1 parameter, 0 given
Its for $insert_stmt->execute()
I don’t know why I have this error, because database and table is true and all of 4 values have sting
if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password,salt) VALUES (?, ?, ?, ?)")) { $insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt); // Execute the prepared query. if (! $insert_stmt->execute()) { echo mysqli_stmt_error(); }
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Answer
The function needs the statement also.
mysqli_stmt_error($yourstatement).
Check it here – mysqli_stmt_error