php - JSON 输出为空

Question

<?php

/*
 * Following code will get single product details
 * A product is identified by product id (pid)
 */

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

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

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

    // get a product from products table
 $result = mysql_query("SELECT * FROM news WHERE Menu = 'FirstPage'");

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

    while ($row = mysql_fetch_array($result)) {
        // temp user array
            $product = array();
            $product["Id"] = $result["Id"];
            $product["Menu"] = $result["Menu"];
            $product["SubPage"] = $result["SubPage"];
            $product["PageName"] = $result["PageName"];
            $product["NewsTitle"] = $result["NewsTitle"];
            $product["PageContent"] = $result["PageContent"];
            $product["Image"] = $result["Image"];
            $product["MenuType"] = $result["MenuType"];
            $product["Date"] = $result["Date"];

            // success
       //     $response["success"] = 1;

            // user node
            $response["news"] = array();

                  array_push($response["news"], $product);
    }
    // success
 //   $response["success"] = 1;

    // echoing JSON response
    echo json_encode($response);
}
?>

JSON 验证器显示

{
    "news": [
        {
            "Id": null,
            "Menu": null,
            "SubPage": null,
            "PageName": null,
            "NewsTitle": null,
            "PageContent": null,
            "Image": null,
            "MenuType": null,
            "Date": null
        }
    ]
}

我正在使用它从 mysql 数据库中获取孟加拉语新闻，并通过 json 脚本为 android 进行解析。你能指导我为什么在我的数据库凭证、参数完全正确的地方得到空值吗？

感谢任何善意的帮助，

提前致谢伊斯提亚克

Answer 1

您为结果集使用了错误的变量。改变这个

while ($row = mysql_fetch_array($result)) {
// temp user array
   $product = array();
   $product["Id"]         = $row ["Id"];
   $product["Menu"]       = $row ["Menu"];
   $product["SubPage"]    = $row ["SubPage"];
   $product["PageName"]   = $row ["PageName"];
   $product["NewsTitle"]  = $row ["NewsTitle"];
   $product["PageContent"] = $row ["PageContent"];
   $product["Image"]      = $row ["Image"];
   $product["MenuType"]   = $row ["MenuType"];
   $product["Date"]       = $row ["Date"];

   // Also you were resetting the $response['news'] 
   // array here unnecessarily
   $response["news"][] = $product;
}

事实上，您可以通过以下方式大大减少此代码

$result = mysql_query("SELECT Id,Menu,SubPage,PageName,NewsTitle,
                              PageContent,Image,MenuType,Date
                      FROM news 
                      WHERE Menu = 'FirstPage'");

if (mysql_num_rows($result) > 0) {
    $response["news"] = array();
    while ($row = mysql_fetch_array($result)) {
        $response["news"][] = $row;
    }
    echo json_encode($response);
}

您还应该考虑使用mysqli_ 或pdo mysql 数据库扩展而不是mysql_ 作为 mysql_ 扩展已弃用，在 PHP7 中将永远消失。