JSON多维使用mysql_fetch_assoc

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我的mysqli更新后的查询，并将其输出使用上面的查询此JSON多维使用mysql_fetch_assoc

SELECT milestone.id, milestone.name, milestone.date, milestone.location, milestone.story_body, milestone.short_link, milestone.created_at, 
GROUP_CONCAT(images.path) as path , images.update_type 
FROM milestone 
INNER JOIN images ON milestone.id = images.update_id 
WHERE milestone.business_id = '1' && milestone.status = '1' && images.update_type = '3' 
GROUP BY milestone.id

我如何形成的JSON对象？

我已经试过以下方法，不给任何结果

$result_array = array(); 
while($row = mysql_fetch_assoc($result)) 
{ 
    $result_array[] = $row; 

}

我想是这样的 -

[ 
     { 
      "id":"4", 
      "name":"2nd anniversary", 
      "date":"2015-12-17", 
      "location":"Mumbai", 
      "story_body":"Gzjjs jdk djks jdks jdkd jx djdb djd JD djbd djdj d", 
      "short_link":"izWfs", 
      "created_at":"2015-12-11 03:49:52", 
      "path": 
       [      
        {"\/SupportData\/ImpalzB2B\/uploads\/90294930451448437444826.jpg"}, 
        {"\/SupportData\/ImpalzB2B\/uploads\/90294930451449758248579.jpg"} 
       ], 
      "update_type":"3" 
     },  
     { 
      "id":"7", 
      "name":"#1styearAnniversary", 
      "date":"2016-01-20", 
      "location":"Mumbai", 
      "story_body":"Bsjsj jdkdk djdkdk dkdkf kdkf dkfj fjfj fjfkjdd djkd", 
      "short_link":"FHXh0", 
      "created_at":"2016-01-20 23:10:54", 
      "path":"\/SupportData\/ImpalzB2B\/uploads\/11453356652175.jpg", 
      "update_type":"3" 
     } 
    ]

注：我知道MySQL是不是在PHP 7中使用。我需要用PDO &替换它，所以请忽略这个错误。与此同时，我正在处理这个问题。

来源

2016-02-05 Nehil Mistry

更新我的查询。 @ A-2-A –

'GROUP_CONCAT（path）'和'GROUP BY id'可以提供帮助。 –

@Sougata我已经更新了我的查询和结果，现在怎样才能得到结果？ –

现在你可以做到这一点 -

$result_array = array(); 
while($row = mysql_fetch_assoc($result)) 
{ 
    $temp= explode(',', $row['path']); // explode by , 
    if(count($temp) > 1) { // More than one element then assign 
     $row['path']= $temp; 
    } 
    $result_array[] = $row; 
}

来源

2016-02-05 08:34:39

JSON多维使用mysql_fetch_assoc

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