從數據庫獲取數據顯示為json響應

Question

在mysql表中我有一組記錄。 我想把它們想要在json響應下面顯示出來。

"results":[

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

我從數據庫獲得了上面的值，即時間戳，id，類別，用戶名。 如何以上面的json響應形式顯示結果？

更新：

我以這種方式獲取數據：

$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

以傳統方式獲取結果：

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

然后將其全部轉換為JSON。 巴姆！

$results = json_encode($data);

這比在SQL查詢中嘗試格式化JSON要容易得多。

有關詳細信息，請參閱

• http://php.net/manual/en/pdostatement.fetchall.php
• http://php.net/manual/en/function.json-encode.php

---

由於您使用mysqli而不是PDO，並以過程方式使用它，因此您將以不同的方式獲取行：

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

然后你可以像我上面所示的json_encode（）。

Answer 2

嘗試這個...

        $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);             
        }