本文介绍了处理 Instagram API (JSON + PHP)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我是 PHP 新手,我不知道如何处理 Instagram API,以便(例如)通过 user_id 3 发布的最近 3 个项目提取标准分辨率图像的链接列表.

这是我迄今为止创建的内容:

';foreach ($data as $key => $value) {$result .='
  • link.'><img src="'.$value->images->standard_resolution->url.'" width="70" height="70"/></a></li>';}$result .= '</ul>';返回 $result;}
  • 结果是一个空白页..你能帮我吗?

    您需要对返回的数据进行回显或执行某些操作(而且您的 json_decode 函数前面还有一个胭脂 $)

    试试这个:

    '.PHP_EOL;foreach ($jsonData->data as $key=>$value) {$result .= "\t".'<a class="fancybox" data-fancybox-group="gallery"title="'.htmlentities($value->caption->text).''.htmlentities(date("F j, Y, g:i a", $value->caption->created_time)).'"style="padding:3px" href="'.$value->images->standard_resolution->url.'"><img src="'.$value->images->low_resolution->url.'" alt="'.$value->caption->text.'" width="'.$width.'" height="'.$height.'"/></a>'.PHP_EOL;}$result .= '</div>'.PHP_EOL;返回 $result;}回声 get_instagram();?>

    使用 $value->location->name

    如果你想检查它是空的,如果它是空的,那么就不要显示它,但如果它被设置,你还想在它上面附加一个字符串,你会做这样的事情:

    $location = (!empty($value->location->name))?'@'.$value->location->name:null;

    然后使用 $location 回显您想要的位置.

    I'm new to PHP and I can't figure out how to deal with Instagram APIs in order to (for example) extract a list of links to the standard resolution images by recent 3 items published by the user_id 3.

    Here's what I created till now:

    <?php
    function get_instagram($user_id,$count)
    {
        $user_id = '3';
        $count = '3';
        $url = 'https://api.instagram.com/v1/users/'.$user_id.'/media/recent/?access_token=13137.f59def8.1a759775695548999504c219ce7b2ecf&count='.$count;
        $jsonData = $json_decode((file_get_contents($url)));
        $data = $jsonData->data;
        $result = '<ul>';
        foreach ($data as $key => $value) {
            $result .= '<li><a href='.$value->link.' ><img src="'.$value->images->standard_resolution->url.'" width="70" height="70" /></a></li> ';
        }
        $result .= '</ul>';
        return $result;
    }
    

    The result is a blank page though.. can you help me?

    You need to echo or do something with the returned data (Also you have a rouge $ in front of your json_decode function)

    Try this:

    <?php
    function get_instagram($user_id=15203338,$count=6,$width=190,$height=190){
    
        $url = 'https://api.instagram.com/v1/users/'.$user_id.'/media/recent/?access_token=13137.f59def8.1a759775695548999504c219ce7b2ecf&count='.$count;
    
        // Also Perhaps you should cache the results as the instagram API is slow
        $cache = './'.sha1($url).'.json';
        if(file_exists($cache) && filemtime($cache) > time() - 60*60){
            // If a cache file exists, and it is newer than 1 hour, use it
            $jsonData = json_decode(file_get_contents($cache));
        } else {
            $jsonData = json_decode((file_get_contents($url)));
            file_put_contents($cache,json_encode($jsonData));
        }
    
        $result = '<div id="instagram">'.PHP_EOL;
        foreach ($jsonData->data as $key=>$value) {
            $result .= "\t".'<a class="fancybox" data-fancybox-group="gallery"
                                title="'.htmlentities($value->caption->text).' '.htmlentities(date("F j, Y, g:i a", $value->caption->created_time)).'"
                                style="padding:3px" href="'.$value->images->standard_resolution->url.'">
                              <img src="'.$value->images->low_resolution->url.'" alt="'.$value->caption->text.'" width="'.$width.'" height="'.$height.'" />
                              </a>'.PHP_EOL;
        }
        $result .= '</div>'.PHP_EOL;
        return $result;
    }
    
    echo get_instagram();
    ?>
    


    With $value->location->name

    If you want to check its empty and if it is then dont show it but also want to append a string onto it if it is set you would do something like:

    $location = (!empty($value->location->name))?'@'.$value->location->name:null;
    

    Then use $location to echo where you want it.

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