php文件錯誤中的SQL查詢

Question

我有這段代碼從一個url傳遞一個變量。當我使用$ _GET方法時，它會返回一個沒有找到產品的json，但是當我手動給出$ user_email從url返回的值時，它會返回正確的json！什麼是錯的，我該如何糾正它？謝謝

網址：HTTP：// * ** * ** * ** * * /android_connect/get_all_products.php?user_email=m

<?php 

/* 
* Following code will list all the products 
*/ 

// array for JSON response 
$response = array(); 

$user_email= $_GET['user_email']; 



// include db connect class 
require_once __DIR__ . '/db_connect.php'; 

// connecting to db 
$db = new DB_CONNECT(); 

// get all products from products table 
$test= "SELECT *FROM products WHERE user_email= '" .$user_email. "'"; 


//echo $test; 
$result = mysql_query($test) or die(mysql_error()); 








// check for empty result 
if (mysql_num_rows($result) > 0) { 
    // looping through all results 
    // products node 
    $response["products"] = array(); 

    while ($row = mysql_fetch_array($result)) { 
     // temp user array 
     $product = array(); 
     $product["pid"] = $row["pid"]; 
     $product["firstname"] = $row["firstname"]; 
     $product["lastname"] = $row["lastname"]; 
     $product["email"] = $row["email"]; 
     $product["phone"] = $row["phone"]; 
     $product["address"] = $row["address"]; 
     $product["created_at"] = $row["created_at"]; 
     $product["updated_at"] = $row["updated_at"]; 
     $product["user_email"] = $row["user_email"]; 


     // push single product into final response array 
     array_push($response["products"], $product); 
    } 
    // success 
    $response["success"] = 1; 

    // echoing JSON response 
    echo json_encode($response); 
    } else { 
    // no products found 
$response["success"] = 0; 
$response["message"] = "No products found"; 

    // echo no users JSON 
    echo json_encode($response); 
     } 
?> 

Answer 1

試試這個：

$test= "SELECT * FROM products WHERE user_email = '$user_email'"; 

編輯：

$test= "SELECT * FROM products WHERE user_email = $user_email";