Question

我有一个返回JSON对象的URL，如下所示：

[
    {
        "idIMDB": "tt0111161",
        "ranking": 1,
        "rating": "9.2",
        "title": "The Shawshank Redemption",
        "urlPoster": "http:\/\/ia.media-imdb.com\/images\/M\/MV5BODU4MjU4NjIwNl5BMl5BanBnXkFtZTgwMDU2MjEyMDE@._V1_UX34_CR0,0,34,50_AL_.jpg",
        "year": "1994"
    }
]

网址：http://www.myapifilms.com/imdb/top

我想获取所有urlPoster值并设置在数组元素中，并将数组转换为JSON以便回显它。

我如何通过PHP实现？

Answer 1

你可以这样做：

<?php 
$json_url = "http://www.myapifilms.com/imdb/top";
$json = file_get_contents($json_url);
$data = json_decode($json, TRUE);
echo "<pre>";
print_r($data);
echo "</pre>";
?>

Answer 2

$json = file_get_contents('http://www.myapifilms.com/imdb/top');

$array = json_decode($json);

$urlPoster=array();
foreach ($array as $value) { 
    $urlPoster[]=$value->urlPoster;
}

print_r($urlPoster);

Answer 3

您可以简单地解码json，然后选择您需要的任何内容：

<?php
$input = '[
        {
                "idIMDB": "tt0111161",
                "ranking": 1,
                "rating": "9.2",
                "title": "The Shawshank Redemption",
                "urlPoster": "http:\/\/ia.media-imdb.com\/images\/M\/MV5BODU4MjU4NjIwNl5BMl5BanBnXkFtZTgwMDU2MjEyMDE@._V1_UX34_CR0,0,34,50_AL_.jpg",
                "year": "1994"
        }


]';

$content = json_decode($input);
$urlPoster = $content[0]->urlPoster;
echo $urlPoster;

输出显然是存储在该属性中的URL：

http://ia.media-imdb.com/images/M/MV5BODU4MjU4NjIwNl5BMl5BanBnXkFtZTgwMDU2MjEyMDE@._V1_UX34_CR0,0,34,50_AL_.jpg

BTW：“肖申克的救赎”是有史以来最好的电影之一......

Answer 4

这是你如何用array_map函数做同样的事情。

<?php

#function to process the input
function process_input($data)
{
 return $data->urlPoster;
}

#input url
$url = 'http://www.myapifilms.com/imdb/top';


#get the data
$json = file_get_contents($url);

#convert to php array
$php_array = json_decode($json);

#process the data and get output
$output = array_map("process_input", $php_array);


#convert the output to json array and print it
echo json_encode($output);

通过PHP从URL获取JSON

4 个答案: