Question

在mysql表中我有一组记录。我想把它们想要在json响应下面显示出来。

"results":[

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

我从数据库获得了上面的值，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

以传统方式获取结果：

$data['results'] = $stmt->fetchAll(PDO::FETCH_ASSOC);

然后将其全部转换为JSON。 BAM！

$results = json_encode($data);

这比在SQL查询中尝试格式化JSON要容易得多。

有关详情，请参阅：

由于您使用mysqli而不是PDO，并以过程方式使用它，因此您将以不同的方式获取行：

while ($data['results'][] = mysql_fetch_assoc($result));

然后你可以像我上面所示的那样json_encode（）。

Answer 2

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);             
        }

从数据库获取数据显示为json响应

2 个答案: