如何將參數發送到另一個PHP網站的AJAX POST方法並獲取JSON信息

Question

我正在研究“ 如何解析其他網站的內容 ”。 我正在使用HTML DOM解析來獲取信息。 我面臨的問題是包含AJAX POST調用以獲取信息的網站。 示例網站： 馬薩諸塞州

1.這里醫生的信息是使用ajax post方法獲取的

Request URL:http://www.massgeneral.org/assets/javascripts/facets/doctors/doctors.ashx
Request Method:POST

我如何在此處傳遞參數以發布方法？ 我試過的是

<?php
echo '<center><h3>Massachusetts Information</h3></center>';
// extra headers
//$headers[]= "Accept-Encoding: gzip, deflate";
$fields['center'] = "";
$fields['centerPreSelected'] = false;
//$fields['displayPaging'] = false;
$fields['gender'] = "";
$fields['isEmpty'] = true;
$fields['languages'] = [];
$fields['letter'] = "A";
$fields['letter'] = "";
$fields['locations'] = array();
$fields['numberOfPages'] = 15;
$fields['numberPerPage'] = 50;
$fields['page'] = 1;
$fields['program'] = "";
$fields['range'] = array('Item1' => 0,'Item2' => 49);
$fields['saytLimit'] = "20";
$fields['term'] = "";

$POSTFIELDS = http_build_query($fields);

$headers[] = "Accept: application/json, text/plain, */*";
$headers[] = "Accept-Encoding: gzip, deflate";
$headers[] = "Accept-Language: en-GB,en;q=0.5";
$headers[] = "Connection: keep-alive";
$headers[] = "Content-Type: application/json";
$headers[] = "Host: www.massgeneral.org";
$headers[]="Referer: http://www.massgeneral.org/doctors/";

$login_submit_url = "http://www.massgeneral.org/assets/javascripts/facets/doctors/doctors.ashx";
$ch = curl_init();
curl_setopt($ch, CURLOPT_HEADER,  0);
curl_setopt($ch, CURLOPT_USERAGENT, $agent);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 1);

curl_setopt($ch, CURLOPT_URL, $login_submit_url);
curl_setopt($ch, CURLOPT_HTTPHEADER,  $headers);
curl_setopt($ch, CURLOPT_VERBOSE, true);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $POSTFIELDS);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 0);
$result = curl_exec($ch);

echo $result;

它沒有獲取醫生的信息。 請提供鏈接或想法進行解析。

Answer 1

您正在使用http_build_query()創建了“ foo = 1＆bar = 2”樣式格式。 該站點需要json，因此您想改用json_encode() 。

另外，除非確定您確實要處理gzip壓縮的響應，否則請忽略gzip標頭。