Question

我已经查看了stackoverflow上的几个旧答案，但它们都已过时，并且它们使用的API已不再可用。

我已经创建了一个JSON / Atom API，CX密钥并使用了一个脚本感谢Adam Fischer我在这里找到但是当我正在尝试我现在能够输出打印结果在页面上我得到了错误：

注意：未定义的属性：stdClass :: $ responseData in   第19行的E：\ XAMPP \ htdocs \ PHP Training \ google.php

注意：尝试在E：\ XAMPP \ htdocs \ PHP中获取非对象的属性   第19行训练\ google.php

这是我到目前为止所拥有的。下面的代码。

$url = 'https://www.googleapis.com/customsearch/v1?key=[MY API KEY]&cx=[MY CX KEY]&q=lecture';

$body = file_get_contents($url);
$json = json_decode($body);

for($x=0;$x<countif ($json->responseData->results);$x++>items){

echo "<b>Result ".($x+1)."</b>";
echo "<br>URL: ";
echoforeach ($json->responseData->results[$x]->url;
echo>items "<br>VisibleURL:as ";$item){
echo $json->responseData->results[$x]->visibleUrl;
echo "<br>Title: ";
echo $json->responseData->results[$x]->title;
echo "<br>Content: ";print_r($item)
echo $json->responseData->results[$x]->content;
echo "<br><br>"; }
}

API工作正常，因为我访问时会吐出数组中的所有内容。示例：dl.dropboxusercontent.com/u/47731225/sample.txt

我正在尝试制作$ url我看到的结果会像Google搜索一样显示在我的页面上，例如：prntscr.com/drum5u

{
   "kind": "customsearch#result",
   "title": "The Tank, Haydon Allen Lecture Theatre, Building 23, ANU",
   "htmlTitle": "The Tank, Haydon Allen \u003cb\u003eLecture\u003c/b\u003e Theatre, Building 23, ANU",
   "link": "https://www.google.com/mymaps/viewer?mid=1YGFZHcZ20jPvy5OiaKT1voy841Q&hl=en",
   "displayLink": "www.google.com",
   "snippet": "\"The Tank\", Haydon Allen Lecture Theatre, Building 23, The Australian National \nUniversity (ANU), Canberra, Australia.",
   "htmlSnippet": "&quot;The Tank&quot;, Haydon Allen \u003cb\u003eLecture\u003c/b\u003e Theatre, Building 23, The Australian National \u003cbr\u003e\nUniversity (ANU), Canberra, Australia.",
   "cacheId": "hTeucZ5TewoJ",
   "formattedUrl": "https://www.google.com/mymaps/viewer?mid...hl=en",
   "htmlFormattedUrl": "https://www.google.com/mymaps/viewer?mid...hl=en",
   "pagemap": {
    "cse_thumbnail": [
     {
      "width": "221",
      "height": "228",
      "src": "https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcSntx5YhQgJQeJ6RAZajOx7SGOwh0oUu8jtpY6VOAS75V_oNkiXx923ro4"
     }

Answer 1

您是否通过API查看了返回的json？我的猜测是，它与你期望的完全不同

见

https://developers.google.com/custom-search/json-api/v1/reference/cse/list

澄清后，您的结果与您的代码所期望的结果完全不同。

正确的代码应该是

$url = 'https://www.googleapis.com/customsearch/v1?key=[MY API KEY]&cx=[MY CX KEY]&q=lecture';

$body = file_get_contents($url);
$json = json_decode($body);
if ($json->items){
   foreach ($json->items as $item){
      print_r($item);
   }
}

Answer 2

您可以使用文件获取内容来获取Google的完整页面内容，并且可以在您的网站中显示结果，例如

function file_get_contents_curl($url) {
    $ch = curl_init();

    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); //Set curl to return the data instead of printing it to the browser.
    curl_setopt($ch, CURLOPT_URL, $url);

    $data = curl_exec($ch);
    curl_close($ch);

    return $data;
}

$query = "search term";
$url = 'http://www.google.co.in/search?q='.urlencode($query).'';
$scrape = file_get_contents_curl($url);

使用PHP输出Google搜索结果？

2 个答案: