Question

嘿，我现在从所有获取信息中都有五个json，遇到这样的问题->从五个不同的json中我需要获取最新 videoId ，谁会先显示对于我来说太辛苦了，将所有功能都放到一个函数foreach中，我尝试将其执行约5小时并保持在同一步骤

这两个json中的所有代码都需要在一个foreach回声中获取最新（最新） videoId

<?php
$videoList1 = json_decode(file_get_contents('https://www.googleapis.com/youtube/v3/search?order=date&part=snippet&channelId=UCKLObxxmmAN4bBXdRtdqEJA&maxResults=50&key=AIzaSyDVTF2abNVa5pRitb8MVz1ceJFhE-2y_qk'));
$videoList2 = json_decode(file_get_contents('https://www.googleapis.com/youtube/v3/search?order=date&part=snippet&channelId=UCynfZM0Edr9cA4pDymb2rEA&maxResults=50&key=AIzaSyDVTF2abNVa5pRitb8MVz1ceJFhE-2y_qk'));

$i = 0;
foreach($videoList1->items as $item){
    if(isset($item->id->videoId)) {  
        echo $item->id->videoId;
        if ( ++$i > 3) {
           break; 
        }
    }
}

Answer 1

纸盒：

$videoList1 = json_decode(file_get_contents('https://www.googleapis.com/youtube/v3/search?order=date&part=snippet&channelId=UCKLObxxmmAN4bBXdRtdqEJA&maxResults=50&key=AIzaSyDVTF2abNVa5pRitb8MVz1ceJFhE-2y_qk'),true);
$videoList2 = json_decode(file_get_contents('https://www.googleapis.com/youtube/v3/search?order=date&part=snippet&channelId=UCynfZM0Edr9cA4pDymb2rEA&maxResults=50&key=AIzaSyDVTF2abNVa5pRitb8MVz1ceJFhE-2y_qk'),true);

$videoList = array_merge($videoList1["items"],$videoList2["items"]);

/// sort lastet first
  foreach ($videoList as $key => $part) {
     $sort[$key] = strtotime($part['snippet']['publishedAt']);
  }
  array_multisort($sort, SORT_DESC, $videoList);

 foreach ($videoList as $video) {
    if(isset($video["id"]["videoId"])) {
      echo 'publishedAt: '. $video['snippet']['publishedAt'] . ' VideoID:  ' .  $video["id"]["videoId"] . "\n </br>";
    } 
 }

一个foreach中有多个json

1 个答案: