mysqli_stmt::bind_param():变量数量与 mysql php 中准备好的语句中的参数数量不匹配

Question

我正在使用准备好的语句并执行

$search = trim($_GET['s']);
if($stmtgetproducts = $mysqli->prepare("SELECT s.store_id, product_id, product_name, product_desc, product_price, product_discount, product_views, product_date, product_code, product_category, product_image1, product_image2, sale_id,s.store_name,s.store_username 
  FROM store_products sp 
  INNER JOIN store_product_categories spc ON sp.product_category=spc.id
  INNER JOIN store s ON sp.store_id=s.store_id WHERE MATCH(product_name, product_desc) AGAINST('?' IN BOOLEAN MODE);")){
$stmtgetproducts->bind_param("s",$search);
$stmtgetproducts->execute();
$getproducts=$stmtgetproducts->get_result();
$stmtgetproducts->close();
}
else
{
 echo $mysqli->error;
} 

因为，我只给它一个参数作为参数，并且绑定(bind)参数只是一个 $search 它给了我这个错误

mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

这个 if 表单

<form action="" method="GET">
                  Search Product: <input type="text" name="s" placeholder="Enter product name">
                </form>

如果我改变这个

$stmtgetproducts->bind_param("ss",$search);

然后就出现这个错误

mysqli_stmt::bind_param(): Number of elements in type definition string doesn't match number of bind variables 

Answer 1

$stmtgetproducts = $mysqli->prepare("SELECT s.store_id, product_id, product_name, product_desc, product_price, product_discount, product_views, product_date, product_code, product_category, product_image1, product_image2, sale_id,s.store_name,s.store_username 
  FROM store_products sp 
  INNER JOIN store_product_categories spc ON sp.product_category=spc.id
  INNER JOIN store s ON sp.store_id=s.store_id WHERE MATCH(product_name, product_desc) AGAINST(? IN BOOLEAN MODE);"))

将我准备好的声明更改为此