php文件中的SQL查询错误

Question

我有这段代码从一个 url 传递一个变量。当我使用 $_GET 方法时，它会返回一个没有找到产品的 json，但是当我手动给出 $user_email 从 url 获得的值时，它会返回正确的 json！出了什么问题，我该如何纠正？谢谢

网址：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

试试这个：

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

编辑：

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