要向访问您网站的访问者隐藏API密钥，请在您自己的网站上使用PHP脚本充当中继。它接收Ajax请求（没有API密钥）；添加您的密钥并发出自己的API请求；然后将响应返回到浏览器。

例如Javascript

var dataString = "item=" + $('#item').val() + "&qty=" + $('#quantity').val(); 
$.ajax({type: "POST", url:"/myrelays/getstockdata.php", data: dataString, success: function(data){ your function to handle returned data } });

getstockdata.php脚本（一个非常粗糙的框架）：

<?php
header('Content-Type: application/json; charset=utf-8');

$api_key = 'xyz1234';
$result = array('status'=>'Error','msg'=>'Invalid parameters');

// your code to sanitize and assign (Ajax) post variables to your PHP variables
// if invalid:   exit(json_encode($result));

// make API request with $api_key
$url = 'https://api.provider.com/stockdata.json?key=' . $api_key . '&item=' . $item . '&qty=' . $qty;
$ch = curl_init($url);  
curl_setopt($ch,CURLOPT_FAILONERROR, TRUE);  // identify as error if http status code >= 400
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,true);  // returns as string
$api_response = curl_exec($ch);
if(curl_errno($ch) || curl_getinfo($ch, CURLINFO_HTTP_CODE) != 200 ) :
    $result['msg'] = 'Item not found or unable to get data. ' . curl_error($ch);
    curl_close($ch);
    exit(json_encode($result));
endif;
curl_close($ch);
$decodedData = json_decode($api_response, true);
// check for success and do any server side manipulation of $decodedData

$result['status'] = 'OK'];
$result['msg'] = '$decodedData';
exit(json_encode($result));
?>

注意：在脚本中，我通常将“ HTML”传递回浏览器。因此，脚本的“ Json”位可能需要更改，例如，可能不需要“ header”（脚本的第一行）。