使用 PHP 从 MySQL 结果输出复杂的 JSON

Question

我刚刚完成了 Sencha Touch 教程，效果很好。我目前正在使用 PHP 将静态 JSON 数据文件（本教程使用）转换为从 MySQL 数据库动态生成的数据 (JSON) 文件。本教程中使用的 JSON 文件（示例文件的链接位于下方）具有复杂的结构，我在将我的 SQL 结果转换为教程中使用的确切 JSON 格式时遇到问题。下面列出了源代码的链接。重点放在 JSON 文件上（第三个链接）。如果我可以简单地复制这个文件，那么项目的其余部分就对我有用。

煎茶触控教程位于： http://www.sencha.com/learn/intro-to-the-nested-list-component/

该项目的所有代码都可以位于（我只允许发布两个链接，希望您能解释下面的链接）： github dot com / nelstrom / Sencha-Touch-nested-list-demo

静态 JSON 数据文件的示例可以查看：https://github.com/senchalearn/Nested-list-demo/blob/master/data/albums.json

我有一个数据库和 sql，它以以下结构输出数据：

Genre  Artist     Album           Track               Duration 
ROCK   MUSE       Absolution      Intro               0:23
ROCK   MUSE       Absolution      Apolcalypse Please  4:13
ROCK   MUSE       The Resistance  Uprising            5:03
ROCK   SEVENDUST  Next            Hero                3:48
FUNK   PRIMUS     Antipop         The Antipop         5:33
FUNK   PRIMUS     Antipop         Ballad of Bodacious 2:29

我有以下输出 JSON 但格式不正确的 php，不幸的是，它已经尽可能接近了 - 抱歉，我的 PHP 有点平均:)

$result = mysql_query($query,$link) or die('Errant query:  '.$query);
$model = array();

if(mysql_num_rows($result)) {
    while($e = mysql_fetch_assoc($result)) {
        $model['Genre'] = $e['Genre'];
        $model['Artist'] = $e['Artist'];
        $model['Album'] = $e['Album'];
        $model['items'][] = array(
                        'text' => $e['Track'],
                        'duration' => $e['Duration']
        );
    }
}  

header('Content-type: application/json');
echo json_encode(array('items'=>$model));}

以下是上面 PHP 代码输出的 JSON 示例：

{
    "items": {
        "Genre": "ROCK",
        "Artist": "MUSE",
        "Album": "Absolution",
        "items": [
            {
                "text": "Intro",
                "duration": "0:23"
            },
            {
                "text": "Apolcalypse Please",
                "duration": "4:13"
            }
        ]
    }
}

很遗憾，这种 JSON 格式不正确。主要问题是循环并在正确的位置应用方括号'['']'。我在下面包含了一个简短的示例：

    {
    "items": [
     {
       "model": "Genre",
       "items": [
         {
           "model": "Artist",
           "items": [
             {
               "model": "Album",
               "items": [
                {
                  "model": "Track",
                  "duration": 96,
                  "text": "Introduction",
                  "items": [

                  ],
                  "info": "",
                  "leaf": true
                 },
                 {
                  "model": "Track",
                  "duration": 155,
                  "text": "Please Accept My Love",
                  "items": [

                  ],
                  "info": "",
                  "leaf": true
                }
              ],
              "text": "Live in Cook County Jail",
              "info": "<p>Live in Cook County Jail is a 1971 live album by B.B. King recorded in Cook County Jail, Chicago, Illinois. It was ranked as number 499 in the book version of Rolling Stone's 500 Greatest Albums of All Time.</p>",
              "leaf": true
            }
          ],
          "text": "B.B.King",
          "info": "<p>Riley B. King aka B. B. King (born September 16th, 1925 in Itta Bena, Mississippi) is a well known American blues guitarist and songwriter. He is among the most respected electric guitarists. </p><p>One of King’s trademarks is naming his guitar (Gibson ES335) “Lucilleâ€. In the 1950s in a bar in Twist, Arkansas two men got into a fight, accidentally knocking over a bucket of burning kerosene (used for heating) and setting the establishment on fire. Risking his life, B.B. King ran back into the collapsing building to retrieve his guitar.</p>",
          "leaf": false
        },
      ],
      "text": "Blues",
      "info": "",
      "leaf": false
    }
  ]
} 

提前感谢您查看此内容，如果内容冗长，我很抱歉，但我只是想确保我已包含所有内容。如果您需要更多信息，请告诉我。

亲切的问候

Answer 1

我要做的第一件事是尝试重新创建 JSON 格式的数据。快速测试

$array = array(
'items' => array(
    array(
    'model' => 'Genre',
    'items' => array(
        array(
        'model' => 'Artist',
        'items' => array(
            'model' => 'Album',
            'items' => array(array())
        )
        )
    )
    )
)
);
$json = json_encode( $array );
var_dump( $json );

这会提供类似于您需要的 JSON 输出。

然后，您需要将数据库中的数据转换为相关格式的数组。为避免多次通过您正在构建的数组（随着数组变大，这可能需要一段时间），我会将此作为一个两步过程来完成。

第一

$tempData = array();
while($e = mysql_fetch_assoc($result)) {
    $tempData[$e['Genre']][$e['Artist']][$e['Album']][$e['Track']] = $e['Duration']
}

然后您应该能够遍历 $tempArray 并以正确的格式构建一个数组来创建 JSON。

Answer 2

您可以使用下面的代码一次性完成此操作。只需确保返回的数据按流派、艺术家、专辑、曲目排序即可。

if(mysql_num_rows($result)) {
   $curr_genre = $curr_artist = $curr_album = $curr_track = '';
   $model = $track = $album = $artist = $genre = array();
   $first_row = 1;
   while ($e = mysql_fetch_assoc($result))
        // every time track changes, create new array and insert into album
        $track['model'] = 'Track';
        $track['info'] = "";
        $track['text'] = $e['Track'];
        $track['duration'] = $e['Duration'];
        $track['leaf'] = TRUE;
        $album['items'][] = $track;
        $track = array();

        // every time album changes, create new array and insert into artist
    if ($curr_album != $e['Album']) {
        $album['model'] = 'Album';
        $album['info'] = "";
        $album['text'] = $curr_album;
        $album['leaf'] = FALSE;
        if (!$first_row) 
    {
        // pop the last item as it's taken one from the next album
        $new_item = array_pop(&$album['items']);
        $artist['items'][] = $album;
        $album = array();
        $album['items'][] = $new_item;
        }
        $curr_album = $e['Album'];

        }

        // every time artist changes, create new array and insert into genre
    if ($curr_artist != $e['Artist']) {
        $artist['model'] = 'Artist';
        $artist['info'] = "";
        $artist['text'] = $curr_artist;
        $artist['leaf'] = FALSE;
        if (!$first_row) 
    {
        $genre["items"][] = $artist;
    $artist = array();
    }
    $curr_artist = $e['Artist'];
    }


        // every time genre changes, create new array and insert into model
    if ($curr_genre != $e['Genre']) {
        $genre['model'] = 'Genre';
        $genre['text'] = $curr_genre;
        $genre['left'] = FALSE;
        if (!$first_row) 
    {
        $model["items"][] = $genre;
    $genre = array();
    }

    $curr_genre = $e['Genre'];
        }

        if ($first_row) {
            $first_row = 0;
        }
    }
  }  
    // now just process the last row

        $album['model'] = 'Album';
        $album['info'] = "";
        $album['text'] = $curr_album;
        $album['leaf'] = FALSE;
        $album['items'][] = $track;

        $artist['model'] = 'Artist';
        $artist['info'] = "";
        $artist['text'] = $curr_artist;
        $artist['leaf'] = FALSE;
        $artist['items'][] = $album;

        $genre['model'] = 'Genre';
        $genre['text'] = $curr_genre;
        $genre['left'] = FALSE;
        $genre["items"][] = $artist;

        $model['items'][] = $genre;

Answer 3

这很棘手。这是一个非常糟糕的数据模型。

您需要添加一些字段，并且可能需要做一些工作，但试试这个。

if(mysql_num_rows($result)) {
    while($e = mysql_fetch_assoc($result)) {
        $genreExisted = 0;
        $albumExisted = 0;
        foreach ($model['items'] as $genreKey => $genre) {
            if ($e['Genre'] == $genre['text']) {
                $genreExisted = 1;
                foreach ($genre['items'] as $albumKey => $album) {
                    if ($e['Album'] == $album['text']) {
                        $albumExisted = 1;
                        //add a new track
                        $model['items'][$genreKey]['items'][$albumKey]['items']["model"] = $e['Track'];
                        $model['items'][$genreKey]['items'][$albumKey]['items']["duration"] = $e['Duration'];
                    }
                }
                if ($albumExisted != 1) {
                    //add an album inside the genre
                    $model['items'][$genreKey]['items'][]["model"] = "Album";
                    $model['items'][$genreKey]['items'][]["text"] = $e['Album'];
                }
            }    
        }
        if ($genreExisted != 1) {
            //add a new genre
            $model['items'][]["model"] = "Genre";
            $model['items'][]["text"] = "Blues";
        }
    }
}