Question

用户有物品。每个物品都有很多属性（价格，购买日期，条件，名称）

我想返回所有项目及其属性的json数组

这是我的代码：

$user_id = $_GET['user_id'];


$query_user_items        = "SELECT product_id FROM user_products WHERE user_id = :user_id";

$item_ids_array = array();
$success = false;

try {
    $sth = $connection->prepare($query_user_items);
    $sth->execute(array(':user_id' => $user_id));   
    $result = $sth->fetchAll(PDO::FETCH_ASSOC);

    $success = true;

} catch (PDOException $ex) {

    $response["success"] = $http_response_server_error;
    $response["message"] = $http_message_server_error . " " . $ex;
    die(json_encode($response));
    $connection = null;
}

if ($success) {

    foreach($result as $key=>$value){

        $query_get_item_details = "SELECT * FROM products WHERE product_id = :product_id";

        $sth = $connection->prepare($qerty_get_item_details);
        $sth->execute(array(':product_id'=> $value));
        $record = $sth->fetch(PDO::FETCH_ASSOC);

        $item_id                = $record['product_id'];
        $item_name              = $record['product_name'];
        $item_time_added        = $record['product_time_added'];
        $item_description       = $record['product_description'];
        $item_brand             = $record['product_brand'];
        $item_price_aquired     = $record['product_price_aquired'];
        $item_bought_from_place = $record['product_bought_from_place'];

    }

    /*
    $response["success"]                        = $http_response_success;
    $response["item_ids_array"]                 = $new_array;

    echo json_encode($response);

我对php不是很好，我不知道在第二次迭代会发生什么（我不知道我是否也正在迭代）

我想$ item_name将被第二项的名称覆盖，然后是第三项，最后我将只有一个对象？如何创建一个对象及其信息数组，然后json_encode呢？

另外，我应该在foreach之外声明这些变量吗？

Answer 1

您的代码可以缩短如下（这样，您不使用任何循环）：

$query = "SELECT p.* FROM user_products u LEFT JOIN products p ON (p.product_id = u.product_id) WHERE u.user_id = :user_id";

try 
{
    $sth = $connection->prepare($query);
    $sth->execute(array(':user_id' => $_GET['user_id']));   
    $response = $sth->fetchAll(PDO::FETCH_ASSOC);
} 
catch (PDOException $ex) 
{
    $response["success"] = $http_response_server_error;
    $response["message"] = $http_message_server_error . " " . $ex;
}
finally
{
    $connection = null;
    echo json_encode($response);
}

然后，在您的ajax成功处理程序中，您可以检查是否设置了成功或消息（如果是这种情况，则会发生错误）。否则，它是一个结果数组，你将遍历它

Answer 2

您不希望每次迭代都运行新查询。您可能希望使用单个查询来获取所有数据：

SELECT p.*
FROM   user_products AS up
LEFT JOIN products AS p ON p.product_id = up.product_id

WHERE up.user_id = :user_id

在PHP中，您可以将数据收集到数组：

<?php
$data = array();
while($result = $sth->fetch(PDO::FETCH_ASSOC)){
        $data[] = $result;
    }

?><?=json_encode($data)?>

Php + MySQL从数据库读取结果并返回与JSON Array类似的对象数组

2 个答案: