[英]How to get JSON in response
通過Chrome瀏覽器調試網站時,會收到JSON響應。 但是,當我嘗試通過PHP執行此操作時,我收到一條錯誤消息。
無法打開流:HTTP請求失敗! 找不到HTTP / 1.0 404
謝謝你的幫助。
例如:
Chrome的活動:
轉到頁面: http ://gruper.pl/warszawa,在底部,您將看到一個按鈕“ Wiecej ofert”。 單擊后,您將在調試中看到:
http://gruper.pl/DataProvider.php?cityId=51&categoryId=0&mainNaviId=1&showBTile=true&page=1
和回應:
[{"ID_PAGE":"59199","ID_CITY":"3952","main_city":"3952","date_start":"2014-02-23 18:00:00","date_end":"2014-03-01 23:59:00","price".....
有可能在PHP中獲得相同的結果嗎?
我的代碼是:
<?php
$url = 'http://gruper.pl/DataProvider.php?cityId=51&categoryId=0&mainNaviId=1&showBTile=true&page=1';
// use key 'http' even if you send the request to https://...
$options = array(
'http' => array(
'header' => "Content-type: application/x-www-form-urlencoded\r\n" .
"Accept:application/json\r\n" .
"Accept-Encoding:gzip,deflate,sdch\r\n" .
"X-Requested-With:XMLHttpRequest\r\n",
'method' => 'GET'
),
);
$context = stream_context_create($options);
$result = (file_get_contents($url, false, $context));
?>
<html>
<head>
<meta charset="UTF-8">
</head>
</html>
除非設置了以下標頭,否則該URL看起來將返回404 HTTP狀態代碼:
X-Requested-With: XMLHttpRequest
Referer: http://gruper.pl/warszawa
所以這將工作:
<?php
$url = 'http://gruper.pl/DataProvider.php?cityId=51&categoryId=0&mainNaviId=1&showBTile=true&page=1';
// use key 'http' even if you send the request to https://...
$options = array(
'http' => array(
'header' => "X-Requested-With: XMLHttpRequest\r\n" .
"Referer: http://gruper.pl/warszawa"
)
);
$context = stream_context_create($options);
$result = (file_get_contents($url, false, $context));
echo $result;
?>
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