mysqli语句出错，选择并更新

Question

您好我在使用select语句时遇到问题，mysqli_prepare返回一个布尔值。我得到了回复＃34; mysqli_stmt_bind_param（）期望参数1为mysqli_stmt，boolean given＆＃34;  我有一个类似的代码，所以我无法找到我的错误。这是我的代码：

$mailUser = $_POST["mailUser"];

$statement = mysqli_prepare($con, "SELECT store.name, product.name, location.hall, location.price FROM location INNER JOIN product ON product.id = location.idProduct INNER JOIN store ON store.id=location.idStore WHERE mailUser= ?");

mysqli_stmt_bind_param($statement, "s", $mailUser);

mysqli_stmt_execute($statement);

另外，我正在尝试使用php变量更新值，它没有显示任何错误，但值没有更新

$statement = mysqli_prepare($con, "UPDATE product SET description='.$description.', brand='.$brand.', altBrand='.$altBrand.', quant='.$quant.', quantToday='.$quantToday.' WHERE name='.$name.' AND mailUser='.$mailUser.'");

mysqli_stmt_execute($statement);

--- --- EDIT 我设法通过将代码更改为此来解决第二个问题：

$ statement = mysqli_prepare（$ con，＆＃34; UPDATE product SET description = ?, brand = ?, altBrand = ?, quant = ?, quantToday =？WHERE mailUser =？and name =？＆＃34;） ;     mysqli_stmt_bind_param（$ statement，＆＃34; sssssss＆＃34;，$ description，$ brand，$ altBrand，$ quant，$ quantToday，$ mailUser，$ name）;     mysqli_stmt_execute（$语句）;