function google_auth_dosearch($string)
{
$request = trim("http://ajax.googleapis.com/ajax/services/search/web");
$referrer = trim("http://localhost/");
//$results = 5;
$version = "1.0";
//$output = 'php';
//$type='phrase';
// entire phrase, limit to 4 results
$getargs =
'?v='.$version.
'&key=ABQIAAAA4oH5MwaexHdhZg4UWRNB1RT2yXp_ZAY8_ufC3CFXhHIE1NvwkxTzUf4N43torAasiY6JD5CaJS6n7Q&rsz=small&q="'.urlencode
($string).'"';
echo $request.$getargs; // Get the curl
session object $session = curl_init($request.$getargs); // Set the GET options.
curl_setopt($session, CURLOPT_USERAGENT, 'Mozilla/4.0 (compatible; MSIE 6.0; Windows NT
5.1');
curl_setopt($session, CURLOPT_HTTPGET, true);
curl_setopt($session, CURLOPT_HEADER, true);
curl_setopt($session, CURLOPT_RETURNTRANSFER, true);
$response = null; // Do the POST and then close the session
$response = curl_exec($session);
curl_close($session);
var_dump($response); // Get HTTP Status code from the response
$status_code = array();
preg_match('/\d\d\d/', $response, $status_code);
echo "<br>";
var_dump(json_decode($response, true));
echo "<br>"; // Check for errors
switch ($status_code[0]) {
case 200:
echo "Success";
break;
case 503:
return
('Your call to Google Web Services failed and returned an HTTP
status of 503. That means: Service
unavailable. An internal problem
prevented us from returning data to
you.');
//break;
case 403:
return
('Your call to Google Web Services failed and returned an HTTP
status of 403. That means: Forbidden.
You do not have permission to access
this resource, or are over your rate
limit.');
//break;
case 400:
// You may want to fall through here and read the specific XML error
return
('Your call to Google Web Services failed and returned an HTTP
status of 400. That means: Bad
request. The parameters passed to the
service did not match as expected. The
exact error is returned in the XML
response.');
//break;
default:
return ('Your call to Google Web Services returned an unexpected HTTP
status of:'.$status_code
[0]);
}
// Get the serialized PHP array from the response, bypassing the header
if (!(preg_match('^{"response.*}^', $response, $json))) {
$json = null;
print_r($response);
}
$results = json_decode($json[0]);
$urls = array();
// for now let's not suport highlighting by attaching query string as
// it will break our unique call later
if (!$results->responseData->results) {
$message = "Invalid \n".$request;
print_r($response);
die($message); }
foreach($results->responseData->results as $result) {
if (substr($result->url, -4) == ".pdf")
$urls[] = $result->cacheUrl;
else
$urls[] = $result->url;
}
return $urls;
}
当我运行这个函数时，我得到：
SuccessHTTP/1.1 200 OK
Pragma: no-cache Expires: Fri, 01 Jan 1990 00:00:00 GMT
Date: Sat, 02 Apr 2011 09:23:27 GMT
Cache-Control: max-age=0, must-revalidate
Content-Type: text/javascript; charset=utf-8
X-Embedded-Status: 200
X-Content-Type-Options: nosniff
X-Frame-Options: SAMEORIGIN X-XSS-Protection: 1; mode=block Server: GSE
Transfer-Encoding: chunked
{"responseData": {"results":[],"cursor":{"moreResultsUrl":"http://www.google.com/search?oe\u003dutf8\u0026ie\u003dutf8\u0026source\u003duds\u0026start\u003d0\u0026hl\u003den\u0026q\u003d%22What+is+the+best+tool+to+do+plagiarism+checking?%22"}}, "responseDetails": null, "responseStatus": 200}
Invalid http://ajax.googleapis.com/ajax/services/search/web
但是即使在浏览器的地址栏中键入生成的查询，Google 中的相同查询也会返回结果返回有效结果
http://ajax.googleapis.com/ajax/services/search/web?v=1.0&rsz=small&q=%22What+is+the+best+tool+to+do+plagiarism+checking
但我无法得到任何结果。谁能帮忙解释一下我做错了什么？