php - 从数据库获取数据，将其显示为 json 响应

Question

在 mysql 表中我有一组记录。我想获取它们，想像下面的 json 响应一样显示。

"results":[

    {
        "timestamp":"2014-03-04 17:26:14",
        "id":"440736785698521089",
        "category":"sports",
        "username":"chetan_bhagat",
        "displayname":"Chetan Bhagat"
     }

我从数据库中获取了以上值，即时间戳、ID、类别、用户名。如何以上面的 json 响应形式显示结果？

更新:

我以这种方式获取数据:

$con = mysqli_connect('127.0.0.1', 'root', '', 'mysql');
            if (mysqli_connect_errno())
            {
                echo "Failed to connect to MySQL: " . mysqli_connect_error();
                return;
            }   
            $today = date("Ymd");           

            $result = mysqli_query($con,"SELECT url,img_url,sentiment,title,category from frrole_cateogry_article where category='".$category."' AND today <= '".$today."' AND title != '' AND img_url != '' order by url desc limit 3 ");
            while ($row = @mysqli_fetch_array($result))
            {
                $url = $row['url'];
                $img_url = $row['img_url'];
                $screen_name = $row['screen_name'];
            }

Answer 1

试试这个...

        $con = mysqli_connect('127.0.0.1', 'root', '', 'mysql');
        if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
            return;
        }   
        $today = date("Ymd");           

        $result = mysqli_query($con,"SELECT url,img_url,sentiment,title,category from frrole_cateogry_article where category='".$category."' AND today <= '".$today."' AND title != '' AND img_url != '' order by url desc limit 3 ");
        while ($row = @mysqli_fetch_array($result))
        {
            json_encode($row);             
        }