为什么不从flickr显示中获取图像? [英] Why won't the images that are grabbed from flickr display?

问题描述

我正在学习在项目中使用flickr api的过程中，并且遇到了问题...在flickr检索照片数据的地方遇到了问题(如下所示)

I'm in the proccess of learning the flickr api, while using it in a project, and have got a problem... I have it where flickr retrieves the photo data (shown below)

但是它本身不会显示图像...我认为这是由于它没有使用下面的src=" "信息构建<img />标签的事实，这是我尝试使用的代码. ..

but it doesn't display the images themselves... I think this is due to the fact that it's not building the <img /> tags with the src=" " info below is the code that I'm attempting to use...

(我的class.flickr.php代码文件)

(my class.flickr.php code file)

class Flickr{

    private $flickr_key;
    private $flickr_secret;
    private $format = 'json';

    // Setting up flickr_key and flickr_secret
    public function __construct( $flickr_key ) {

        $this->flickr_key = $flickr_key;
    }

    public function searchPhotos( $query = '', $tags = '' ){ // Begin searchPhotos 

        $urlencoded_tags = array( 'animals', 'design', 'phones');

        if ( !empty( $args )) {

            $tags_r = explode( ',', $tags );
            foreach ( $tags_r as $tag ) {

                $urlencoded_tags[] = urlencode( $tag );
            }
        }

        // Construct the url
        $url  = 'http://api.flickr.com/services/rest/?';
        $url .= 'method=flickr.photos.search';
        $url .= '&text=' . urlencode( $query );
        $url .= '&tags=' . implode( ',', $urlencoded_tags );
        $url .= '&sort=relevance';
        $url .= '&safe_search=1';
        $url .= '&content_type=4';
        $url .= '&api_key=' . $this->flickr_key;
        $url .= '&format' . $this->format;
        $url .= '&per_page=10';

        // Calling url using curl
        $curl = curl_init();

        curl_setopt_array( $curl, array(

            CURLOPT_TIMEOUT => 120,
            CURLOPT_URL => $url,            
        ));

        if ( !curl_exec( $curl )) {

            die ( 'Error: "' . curl_error( $curl ) . '" - Code: ' . curl_errno( $curl ));
        }

        // Get search results
        $result = file_get_contents( $url );

        // Remove the unneccessary strings that wraps the result returned from the API
        $json = substr( $result, strlen( "jsonFlickrApi("), strlen( $result ) - strlen( "jsonFlickrApi(") - 1 );

        $photos = array();
        $data = json_decode( $json, true );

        // Check if the status didn't fail
        if ( $data['stat'] != 'fail' ) {

            /** Return only the data for the photos 
                as that's the only thing that we need
            */
            $photos = $data['photos']['photo'];
            return $photos;
        } else {

            return false;       
        }
    } // end searchPhotos

}

(我在页面模板文件中使用的方法调用)

(my method call which I use in my page template file)

<?php // Flickr search photos test

require_once( 'class.flickr.php' );

global $flickr_key;

$flickr = new Flickr( 'this_is_my_api-key_placeholder...' );

$query = "Event Photo Uploadr";

$results = $flickr->searchPhotos( $query, $tag );
if ( !empty( $results )) {

    foreach( $results as $photo ) {

        $src = "http://farm" . $photo['farm'] . ".static.flickr.com/" . $photo['server'] . '/' . $photo['id'] . '_' . $photo['secret'] . '.jpg';
        ?>

        <img  src="<?php echo $src; ?>"/>

    <?php }                 
}

正如我所说的，我是flickr api的新手...到目前为止，我很幸运能做到这一点(感谢stackoverflow，Wordpress Answers和开发者朋友)，因此，非常感谢任何有用的输入！ ;)

As I said I'm new to flickr api... I've been blessed to make it thus far (thanks to stackoverflow, Wordpress Answers and developer friends), so any useful input is GREATLY appreciated! ;)

推荐答案

好的，这听起来很愚蠢，但是，真正的问题是我错过了=.我不是在开玩笑，我有这个$url .= '&format' . $this->format;(使api，默认情况下返回xml)应该是这个$url .= '&format=' . $this->format;(这是获取我的$format var的正确方法).我还整理了一下代码...

Okay, this is going to sound silly, but, the real problem was I was missing a =. I'm not joking, I had this $url .= '&format' . $this->format; (which made the api, by default return xml) should be this $url .= '&format=' . $this->format; (this is the proper way to grab my $format var). I also cleaned up my code a bit...

(主要的flickr功能文件)

(main flickr functionality file)

<?php

class Flickr{

private $flickr_key;
private $flickr_secret;
private $format = 'json';

// Setting up flickr_key and flickr_secret
public function __construct( $flickr_key ) {

    $this->flickr_key = $flickr_key;
}

public function searchPhotos( $query = '', $tags = '' ){ // Begin searchPhotos 

    $urlencoded_tags = array( 'animals', 'design', 'phones');

    if ( !empty( $args )) {

        $tags_r = explode( ',', $tags );
        foreach ( $tags_r as $tag ) {

            $urlencoded_tags[] = urlencode( $tag );
        }
    }

    // Construct the url
    $url  = 'http://api.flickr.com/services/rest/?';
    $url .= 'method=flickr.photos.search';
    $url .= '&text=' . urlencode( $query );
    $url .= '&tags=' . implode( ',', $urlencoded_tags );
    $url .= '&sort=relevance';
    $url .= '&safe_search=1';
    $url .= '&content_type=4';
    $url .= '&api_key=' . $this->flickr_key;
    $url .= '&format=' . $this->format;
    $url .= '&per_page=10';

    // Get search results
    $result = file_get_contents( $url );

    // Remove the unnecessary strings that wraps the result returned from the API
    $json = substr( $result, strlen( "jsonFlickrApi(" ), strlen( $result ) - strlen( "jsonFlickrApi(" ) - 1 );

    $photos = array();
    $data = json_decode( $json, true );

    // Check if the status didn't fail
    if ( $data['stat'] != 'fail' ) {

        // Return only the data for the photos as that's the only thing that we need
        $photos = $data['photos']['photo'];
        return $photos;

    } else {

        return false;       
    }
} // end searchPhotos

}

我在方法调用中所做的唯一更改是添加了一个alt标记(如alt="<?php echo $photo['title']?>"这样)，并向else语句echo "Failed to construct url successfully";添加了一个错误信息.

The only things that I changed in my method call was adding an alt tag, like so alt="<?php echo $photo['title']?>", and adding a error messege to the else statement, echo "Failed to construct url successfully";.

这篇关于为什么不从flickr显示中获取图像?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！