使用mysql_fetch_assoc的JSON多维数组

问题描述：

My Mysqli Updated Query and it Outputs this

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

How Can I form JSON object using above query?

I've tried below method which doesn't give any result

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

}

I want something like this -

[  
       {  
          "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"
       }
    ]

Note: I know Mysql is not being used in PHP 7. I need to replace it with PDO & Mysqli so please neglect that mistake. I am working on same meanwhile I am facing this query.

我的Mysqli更新查询并输出此 p>

  SELECT里程碑 .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里程碑
INNER JOIN图像ON里程碑。  id = images.update_id 
WHERE milestone.business_id ='1'＆amp;＆amp;  milestone.status ='1'＆amp;＆amp;  images.update_type ='3'
GROUP BY milestone.id 
  code>  pre> 
 
     p> 
 
 
如何使用上述查询形成JSON对象？ p> 
 
 
我尝试过以下方法，但没有给出任何结果 p> 
 
 
  $  result_array = array（）; 
while（$ row = mysql_fetch_assoc（$ result））
 {
 $ result_array [] = $ row; 
 
} 
  code>  pre> 
 
  我想要这样的东西 -   p> 
 
 
  [
 {
“id”：“4”，
“name”：“2周年纪念日”，
  “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”  }，
 {“\ /支持 Data \ / ImpalzB2B \ / uploads \ /90294930451449758248579.jpg“} 
 
]，
”更新_类型“：”3“
}，
 {
”id“：”7“，
”名称“：  “＃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“，  n“update_type”：“3”
} 
] 
  code>  pre> 
 
 注意：我知道PHP 7中没有使用Mysql。我需要将其替换为 PDO＆amp;  Mysqli所以请忽略那个错误。 我正在努力同时我正面临这个问题。   p> 
  div>

答

You can do this now -

$result_array = array();
while($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;
}

使用mysql_fetch_assoc的JSON多维数组

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