Question

php 如何将获取json中的参数的值

json格式{"code":200,"message":"解析成功。","source":"bilibili","sourceName":"哔哩哔哩","vid":"8453162","info":{"title":"【7月】SHOW BY ROCK!! Short!! 01【独家正版】","duration":180,"image":"123456.jpg"},"result":{"duration":180.013,"files":[{"url":"123456.mp4","duration":180.013,"filesize":0,"filetype":"mp4"}],"h":2},"definition":"sd","definitionList":"sd|hd","cost":0.005}
我只想获取 url 里的 123456.mp4 以echo方式写出~ 求大神帮助 谢谢
这是一个URL里获取的json数据aa.php中，我是php小白 求大神给个例子~谢谢

Answer 1

首先 复制你的那条乱乱的json 然后打开

http://www.bejson.com/

丢进去转换 然后替换下面的函数 即可

代码如下

    $object = json_decode('
        {    
           "code": 200,
           "message": "解析成功。",
           "source": "bilibili",
           "sourceName": "哔哩哔哩",
           "vid": "8453162",
           "info": {
               "title": "【7月】SHOW BY ROCK!! Short!! 01【独家正版】",
               "duration": 180,
               "image": "123456.jpg"
           },
           "result": {
               "duration": 180.013,
               "files": [
                   {
                       "url": "123456.mp4",
                       "duration": 180.013,
                       "filesize": 0,
                       "filetype": "mp4"
                   }
               ],
               "h": 2
           },
           "definition": "sd",
           "definitionList": "sd|hd",
           "cost": 0.005
        }
    ');
    foreach($object as $key => $value){
        if(is_object($value)){
            foreach ($value as $k => $v){
                echo $k . ' : ' . $v . '</br>';
            }
        }else{
            echo $key . ' : ' . $value . '</br>';
        }
    }

望采纳~

Answer 2

使用json_decode将文本内容转换成对象或者数组，然后取值

Answer 3

转为数组，for，循环输出