使用PHP输出Google搜索结果？

Question

我看了几个旧的答案在stackoverflow，但他们都过时了，他们使用的API不再可用。

我创建了一个JSON/Atom API，CX键并使用了一个脚本感谢Adam Fischer我在这里找到了，但是当我试着现在能够输出打印页面上的结果时，错误：

Notice: Undefined property: stdClass::$responseData in E:\XAMPP\htdocs\PHP Training\google.php on line 19

Notice: Trying to get property of non-object in E:\XAMPP\htdocs\PHP Training\google.php on line 19

这是我到目前为止。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>"; } 
} 

该API工作正常，因为当我访问时这将吐出数组中的所有内容。例如：dl.dropboxusercontent.com/u/47731225/sample.txt

我想让$ url看到结果，例如在Google我的网页上显示的结果，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": ""The Tank", 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

你去翻从API返回的JSON显示结果？我的猜测是，这是完全不同的，对你的期望是什么

见

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

clarifiacation后，结果与预期你的代码真的不同。

正确的代码看起来应该像

$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

您可以使用该文件获取内容，以获得谷歌的整页内容，您可以在您的网站，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);