php session只能存儲和顯示最近的5個視圖產品

Question

我想做一個會話存儲，然後顯示用戶看到的最後5個產品...但不適合工作

首先存儲相同ID產品，如果我referesh，我想只存儲不同的ID（產品） 2.其次是不顯示我超過1行（1查詢），即使我將有更多的ID商店（不同的產品看到像用戶ID： 3,5,5,6,2）只顯示我在這種情況下的最後一個ID商店 2 ....我想顯示所有的查詢..

。

if (isset($_GET['id'])) { 
    // Connect to the MySQL database 
    include "storescripts/connect_to_mysql.php"; 
    $id = preg_replace('#[^0-9]#i', '', $_GET['id']); 
    $_SESSION['lastViewProducts'][] = $id; 
    if(count ($_SESSION['lastViewProducts']) > 5){ 
     array_shift($_SESSION['lastViewProducts']); 
     } 
    foreach($_SESSION['lastViewProducts'] as $keyLASTview=>$valueLASTview) { 
     $stmtLastVIEW = $con->prepare('SELECT id, product_name, price, details, category, subcategory, size, date_added, image, brand_name, product_color, vizualizari FROM products WHERE id=?'); 
     $stmtLastVIEW->bind_param('i', $valueLASTview); 
     $stmtLastVIEW->execute(); 
     $stmtLastVIEW->bind_result($idSelectDetalii, $produsNumeDetalii, $priceDetalii, $produsDetalii, $produsCategory, $produsSubcategory, $produsSize, $produsDate_added, $imageLocationDetalii, $brandProdusSelectat, $produsColor, $produsVizualizari); 
     $lastVIEWproduct = ''; 
     while ($stmtLastVIEW->fetch()) { 
      $lastVIEWproduct .='titlu: '.$idSelectDetalii.' <img src="'.$imageLocationDetalii.'" class="img-responsive">'; 
      } 
     $stmtLastVIEW->free_result(); 
    } 
} 

我真的卡住了...我無法找出解決這個問題，也許其他變種？ 另外3.第三如果用戶在同一頁上的products.php？id = 3和id = 3它存儲在會話中我不想顯示這個... WHERE id=?(for the showing last 5 products views) and id is not (the $id)?如何編寫mysql select語句和where條款？如果我想也否定

Answer 1

你的邏輯有一些錯誤。您可以使用此代碼嘗試：

if (isset($_GET['id'])) { 

    // Connect to the MySQL database 
    include "storescripts/connect_to_mysql.php"; 
    $id = preg_replace('#[^0-9]#i', '', $_GET['id']); 

    //1. Store only differents ids in the session by creating an associative array 
    $_SESSION['lastViewProducts'][$id] = $id; 

    //3. Don't show the $_GET['id'] in the list 
    $ids = array_filter($_SESSION['lastViewProducts'], function($currentID) use($id) { 
     return $id != $currentId; 
    }); 


    //Keep only 5 distincts articles 
    $ids = array_slice(array_unique($ids), 0, 5); 

    //2. declare your string OUTSIDE the foreach 
    $lastVIEWproduct = ''; 
    foreach($ids as $valueLASTview) { 
     $stmtLastVIEW = $con->prepare('SELECT id, product_name, price, details, category, subcategory, size, date_added, image, brand_name, product_color, vizualizari FROM products WHERE id=?'); 
     $stmtLastVIEW->bind_param('i', $valueLASTview); 
     $stmtLastVIEW->execute(); 
     $stmtLastVIEW->bind_result($idSelectDetalii, $produsNumeDetalii, $priceDetalii, $produsDetalii, $produsCategory, $produsSubcategory, $produsSize, $produsDate_added, $imageLocationDetalii, $brandProdusSelectat, $produsColor, $produsVizualizari); 
     while ($stmtLastVIEW->fetch()) { 
      $lastVIEWproduct .='titlu: '.$idSelectDetalii.' <img src="'.$imageLocationDetalii.'" class="img-responsive">'; 
     } 

     $stmtLastVIEW->free_result(); 
    } 
}