instagram标签api功能

Question

Can someone help me fix this code? result only appear 1 picture..

function get_instagram($q,$client_id) {
        $api = "https://api.instagram.com/v1/tags/".$q."/media/recent?client_id=".$client_id;
        $response = get_curl($api);
        $images = array();
        if($response){
            foreach(json_decode($response)->data as $item){     
                $src = $item->images->standard_resolution->url;
                $thumb = $item->images->thumbnail->url;
                $url = $item->link;

                $images[] = array(
                "src" => htmlspecialchars($src),
                "thumb" => htmlspecialchars($thumb),
                "url" => htmlspecialchars($url)
                );
                return "<a href='".$url."' target='_blank'><img src='".$thumb."' border='0'/></a>";
            }
        }
}

What I'm trying to achieve is make all result appear in a page

Answer 1

In your code, you are getting only one image because, you are returning image inside loop so it dont executes all the records inside loop. Try below code

function get_instagram($q,$client_id) {
    $api = "https://api.instagram.com/v1/tags/".$q."/media/recent?client_id=".$client_id;
    $response = get_curl($api);
    $images = array();
    $returnval = '';
    if($response){
        foreach(json_decode($response)->data as $item){     
            $src = $item->images->standard_resolution->url;
            $thumb = $item->images->thumbnail->url;
            $url = $item->link;

            $images[] = array(
            "src" => htmlspecialchars($src),
            "thumb" => htmlspecialchars($thumb),
            "url" => htmlspecialchars($url)
            );
            $returnval .= "<a href='".$url."' target='_blank'><img src='".$thumb."' border='0'/></a>";
        }
    }
    return $returnval;
}