PHP 从远程 URL 读取 JSON 文件

Question

我遇到了这个问题，PHP 和 Laravel。我正在尝试从远程 ULR 读取 JSON 文件:

https://services.realestate.com.au/services/listings/search?query={"channel":"buy","filters":{"propertyType":["house"],"surroundingSuburbs":"False","excludeTier2":"true","geoPrecision":"address","localities":[{"searchLocation":"Blacktown, NSW 2148"}]},"pageSize":"100"}

我使用的代码:

$re_url = 'https://services.realestate.com.au/services/listings/search?query={"channel":"buy","filters":{"propertyType":["house"],"surroundingSuburbs":"False","excludeTier2":"true","geoPrecision":"address","localities":[{"searchLocation":"Blacktown, NSW 2148"}]},"pageSize":"100"}';

$ch = curl_init($re_url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$re_str = curl_exec($ch);
curl_close($ch);
$re_list = json_decode($re_str);

它不断收到错误“处理您的请求时发生错误。引用#30.96464868.1492255689.1829cf2”

我尝试了“https://google.com.au”的url，效果很好，所以看起来像是URL编码问题。但我不确定。

有人可以帮忙吗，或者有同样的问题吗？

谢谢

Answer 1

我相信您有两个问题，下面的代码可以解决它们。我的代码还使用了一些不同的方法来避免手动组装 JSON、URL 查询字符串等(请参阅所提供代码的第 3-41 行)

问题

1. 您没有对查询参数值进行编码 - 可以使用参数值的 urlencode 进行修复，但我更喜欢 http_build_query，原因已在我的介绍性段落中指出。<
2. 您没有发送用户代理 (UA) header (远端似乎需要此 header 中的值，但不关心它是什么。收到带有 UA 的请求后，我认为它必须将 IP 列入白名单一会儿，因为它似乎并不需要在每个请求上都需要它。不过，我只会为每个请求发送它，因为它不会造成伤害，而且你永远不知道你的白名单何时会超时)。请参阅第 50-53 行，了解我在此脚本中设置的内容以及您拥有的一些选项

替换代码

带有解释性注释

<?php

/*
 * The data that will be serialized as JSON and used as the value of the
 * `query` parameter in your URL query string
 */
$search_query_data = [
    "channel" => "buy",
    "filters" => [
        "propertyType" => [
            "house",
        ],
        "surroundingSuburbs" => "False",
        "excludeTier2" => "true",
        "geoPrecision" => "address",
        "localities" => [
            [
                "searchLocation" => "Blacktown, NSW 2148",
            ],
        ],
    ],
    "pageSize" => "100",
];

/*
 * Serialize the data as JSON
 */
$search_query_json = json_encode($search_query_data);

/*
 * Make a URL query string with a param named `query` that will be set as the
 * JSON from above
 */
$url_query_string = http_build_query([
    'query' => $search_query_json,
]);

/*
 * Assemble the URL to which we'll make the request, and set it into CURL
 */
$request_url = 'https://services.realestate.com.au/services/listings/search?' . $url_query_string;

$ch = curl_init($request_url);

/*
 * Set some CURL options
 */
// Have `curl_exec()` return the transfer as a string instead of outputting
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
// Set a user agent header
curl_setopt($ch, CURLOPT_USERAGENT, 'H.H\'s PHP CURL script');
// If you want to spoof, say, Safari instead, remove the last line and uncomment the next:
//curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_4) AppleWebKit/603.1.30 (KHTML, like Gecko) Version/10.1 Safari/603.1.30');

/*
 * Get the response and close out the CURL handle
 */
$response_body = curl_exec($ch);
curl_close($ch);

/*
 * Unserialize the response body JSON
 */
$search_results = json_decode($response_body);

最后，顺便说一句，我建议您停止直接使用 CURL，并开始使用库来抽象一些 HTTP 交互，并使您的请求/响应开始更好地适应“标准”(PSR) 接口(interface)。由于您使用的是 Laravel，因此您已经处于具有 Composer 的生态系统中，因此您可以轻松 install something like Guzzle .