如何将 json 数据转换为逗号分隔的 php 字符串？答案

【问题标题】：How to transform json data to a comma separated php string?如何将 json 数据转换为逗号分隔的 php 字符串？
【发布时间】：2012-04-05 06:43:24
【问题描述】：

假设我有这个 json 数据。如何将“标签”转换为字符串，如

$tags = "Rihanna, We, Found, Love, (Explicit), Def, Jam, Records, Pop";?

{ "apiVersion" : "2.1",
      "data" : { "items" : [ { "accessControl" : { "autoPlay" : "allowed",
                    "comment" : "allowed",
                    "commentVote" : "allowed",
                    "embed" : "allowed",
                    "list" : "allowed",
                    "rate" : "allowed",
                    "syndicate" : "allowed",
                    "videoRespond" : "allowed"
                  },
                "aspectRatio" : "widescreen",
                "category" : "Music",
                "tags" : [ "Rihanna",
                    "We",
                    "Found",
                    "Love",
                    "(Explicit)",
                    "Def",
                    "Jam",
                    "Records",
                    "Pop"
                  ],
                "title" : "Rihanna - We Found Love ft. Calvin Harris"
              } ],
          "itemsPerPage" : 1,
          "startIndex" : 1,
          "totalItems" : 859012,
          "updated" : "2012-04-04T20:32:26.170Z"
        }
    }

以标题为例，脚本如下：

$content = $this->getDataFromUrl($feedURL);
$content = json_decode($content,true);

$videosList = $content['data']['items'];

for($i=0; $i<count($videosList); $i++) {

$videosDatas['videos'][$i]['title'] = $videosList[$i]['title'];

}

【问题讨论】：

你想把标签写到$videosDatas['videos'][$i]['tags']吗？

标签： php json

【解决方案1】：

您似乎需要implode()。该功能将类似于...

$tags = implode(', ', $videosDatas['videos'][$i]['tags']);

【讨论】：

你需要把$tags改成'tags'

【解决方案2】：

$comma_sep_string = implode(', ' , $videosDatas['videos'][$i]['tags']);

【讨论】：

【解决方案3】：

试试：

foreach($videosList as $i=>$video){
    $videosDatas['videos'][$i]['title'] = $video->title;
    $tags = implode(', ',$video->tags);
    $videosDatas['videos'][$i]['tags'] = $tags;
}

代替您的代码：

for($i=0; $i<count($videosList); $i++) {

$videosDatas['videos'][$i]['title'] = $videosList[$i]['title'];

}

【讨论】：