编辑 API URL 的范围

Question

大家好，我需要一些帮助来编辑基于网站 url 的 API 范围，所以如果我有 foo.com/?Wab_id=15 它将将 API 范围编辑为

?scope=accepted&view=films&wab_id=15

我在想我可以有类似

的东西

$ApiData = file_get_contents('http://foo.com/api/?scope=accepted&view=films/&wab_id=$id');

然后使用 Get 检索传递给 url 的 ID，以编辑 API 的 url。我还尝试循环遍历整个 json 然后调用数组内的 a 键但也没有得到太多运气我的代码在下面

    $ApiData = file_get_contents('http://foo.com/api/?scope=accepted&view=films');
$obj = json_decode($ApiData, true);
$data = $obj;
//here you load $data with whatever you want.
$id = $_GET['id'];
foreach ($data[$id] as $key=>$value){
echo "$key -> $value<br>";
}
?>

但是这个 returns 是

的错误

Invalid argument supplied for foreach()

我还尝试在 foreach 中使用 foreach 循环遍历 Muti 数组，并显示了代码和结果下面的值

$obj = json_decode($ApiData, true);
$data = $obj;
//here you load $data with whatever you want.

foreach ( $data as $film ){
  foreach ( $film as $key=>$val ){
    echo "$key -> $val<br>";
  }
}

结果

uid -> 95
wab_id -> 95
title -> La Batalla de los Invisibles
title_en -> Battle of the Invisibles
syn_sm ->
syn_med -> 
syn_lg -> 
form -> 
genre -> 
language -> 
subtitle_lang -> 
year -> 
runtime -> 
place_sub_city -> 
place_sub_state -> 
place_sub_country -> 
place_film_country -> Mexico
place_dir_city -> 
place_dir_state -> 
place_dir_country -> 
accepted -> 1
festival_year -> 2014
trailer -> 
links ->

Answer 1

正如我在评论中提到的：

Are you passing wab_id, Wab_id, or id to your server where you will access with $_GET? Because its not clear under which key you are passing it to yourself and if you are using the correct one - which may be the issue you are having right now.

在你有了那个平方之后，操纵 API 参数应该是一件简单的事情：

$api = "http://foo.com/api/";
$params = array(
   'scope' => 'accepted',
   'view' => 'films',
);

// you need to match the key you are using here to what you are passing
// to your URL
if (isset($_GET['id'])) {
  $params['wab_id'] = $_GET['id'];
}

$url = $api . '?' . http_build_query($params);

现在对于最后一部分，你真的应该使用 cURL、Http... 或者 file_get_contents 以外的任何其他东西，因为它并没有真正给你一个很好的方法处理您可能遇到的错误。我将使用 cURL:

给你一个简单的例子

$client = curl_init();
curl_setopt_array($client, array(
   CURLOPT_RETURNTRASNFER => true,
   CURLOPT_URL => $url // the one we dynmically built above
));

$responseText = curl_exec($client);

if ($responseText !== false) {
   $responseInfo = curl_getinfo($client);

   if ($responseInfo['http_code'] === 200) {
       // http status ok - you may need to take action based 
       // on other http status codes but im not going to delve into that here.

       $data = json_decode($responseText, true);
       print_r($data);

    } else {
      printf('Error accessing data: HTTP Status %s', $responseInfo['http_code'];
    }
} else {
  printf('Error: (%s) %s', curl_errno($client), curl_error($client));
}

curl_close($client);

编辑 API URL 的范围

Editing the scope of an API URL

php

api

url

foreach