PHP Reforce搜索引擎

Question

我有這個“搜索”選項，它在其中查找“標題”和“類別”，指的是FORM請求的變量。

//This could be Reference or Product Name
$name = mysqli_real_escape_string($con, sanitize($_GET['search']));

//this could be a specific category or Empty(used to reference number)
$category = mysqli_real_escape_string($con, sanitize($_GET['category']));
$x = 0; 
$q = str_replace(array("\\",";"), "", $name);  // remove ALL backslashes & remove ALL ";" -> for sql security: no (simple) injection of commands
$q = trim($q);
$search_exploded = explode(" ", $q);

foreach($search_exploded as $search_each ) { 

$x++; 

   if($x == 1) { 
         $wherearr[]= "ads_title LIKE '%$search_each%' AND category_id = '$category' AND ads_active = 1 AND ads_end = 0"; 
   } else {     
         $wherearr[]= "ads_title LIKE '%$search_each%' AND category_id = '$category' AND ads_active = 1 AND ads_end = 0"; 
   }

}

//$wherearr[] is used to create a new variable $construct to insert to SQL like "Select * from Where $construct"    

我想讓人們也可以通過產品參考進行搜索。

目標是，當該人插入參考時，它將自動轉到該產品。

添加此字段：

 $wherearr[]= "ads_reference LIKE '%$search_each%' AND ads_active = 1 AND ads_end = 0";

如何使用之前創建的表單來完成此操作？

Answer 1

不確定我是否完全理解您的問題，但我會嘗試。

首先嘗試從GET請求中提取所需的所有數據。

一些假設：

1. 您使用AND category_id = '$category'所以我假設$_GET['category']檢索category_id。
2. 提取$_GET['search'] ，將獲得名稱列表，並用空格分隔。 因此，如果您具有foo和bar之類的名稱，則$_GET['search']將為“ foo bar”
3. 如果category_id為空，我認為$name包含引用

現在嘗試以下代碼：

$name = mysqli_real_escape_string($con, sanitize($_GET['search']));
$category_id = mysqli_real_escape_string($con, sanitize($_GET['category']));
$q = str_replace(array("\\",";"), "", $name); 

$search_exploded = explode(" ", trim($q));
$wherearr = array();
$querySufix = " AND ads_active = 1 AND ads_end = 0";

foreach($search_exploded as $search_each) {
    if ($category_id)
        $wherearr[]= "ads_title LIKE '%$search_each%' AND category_id = '$category_id '" . $querySufix;
    else 
        $wherearr[]= "ads_reference LIKE '%$search_each%'". $querySufix;
}