PHP增强搜索引擎

Question

我有这个搜索选项，在这里它查找标题和类别引用到表单请求的变量。

//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"    

我想让有能力的人也可以通过产品参考来搜索。

目标是，当用户插入引用时，它将自动进入产品。

添加此字段：

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

如何使用我以前创建的表单来完成它呢？

Answer 1

我不确定我是否完全理解你的问题，但我会尽力的。

首先，尝试从GET请求中提取您想要的所有数据。

一些假设：

1. 您使用了AND category_id = '$category'，所以我假设$_GET['category']检索category_id。
2. 在提取$_GET['search']时，可以获得按空格分隔的名称列表。因此，如果您有像foo和bar这样的名称，那么$_GET['search']将是"foo bar“。
3. 如果category_id是空的，我假设$name包含引用

现在试试下面的代码：

$name = mysqli_real_escape_string($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;
}