通过file_get_contents通过标签获取推文


Getting tweets by hashtags via file_get_contents

我试图使用Twitter搜索来获取标签的推文。

由于我只需要读取一个json文件,所以我使用了file_get_contents而不是像那样的cURL

file_get_contents('http://search.twitter.com/search.json?q=%23trends');

但是,当我在本地主机中运行代码时,它会显示以下错误。你知道吗?

警告:file_get_contents(http://search.twitter.com/search.json)[function.file-get-contents]:无法打开流:找不到套接字传输"ssl"-配置PHP时忘记启用它了吗

除了FGC是一个本地文件外,在任何情况下都最好使用对FGC的卷曲。

如果未启用extension=php_curl.dll,请在localhost上启用curl(99.9%的活动主机已启用curl。)

试试这个:

<?php 
//Grab the twitter API with curl
$result = curl_get('http://search.twitter.com/search.json?q=%23trends');
//Decode the json result
//into an Object
$twitter = json_decode($result);
//into an Array
$twitter = json_decode($result,true);
//Debug
print_r($twitter);
function curl_get($url){
    $return = '';
    (function_exists('curl_init')) ? '' : die('cURL Must be installed!');
    $curl = curl_init();
    $header[0] = "Accept: text/xml,application/xml,application/json,application/xhtml+xml,";
    $header[0] .= "text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5";
    $header[] = "Cache-Control: max-age=0";
    $header[] = "Connection: keep-alive";
    $header[] = "Keep-Alive: 300";
    $header[] = "Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7";
    $header[] = "Accept-Language: en-us,en;q=0.5";
    curl_setopt($curl, CURLOPT_URL, $url);
    curl_setopt($curl, CURLOPT_USERAGENT, 'TwitterAPI.Grabber (http://example.com/yoursite)');
    curl_setopt($curl, CURLOPT_HTTPHEADER, $header);
    curl_setopt($curl, CURLOPT_HEADER, 0);
    curl_setopt($curl, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($curl, CURLOPT_FOLLOWLOCATION, true);
    curl_setopt($curl, CURLOPT_TIMEOUT, 30);
    curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, false);
    curl_setopt($curl, CURLOPT_HTTPPROXYTUNNEL, true);
    curl_setopt($curl, CURLOPT_PROXY, 'http://proxy_address/');
    curl_setopt($curl, CURLOPT_PROXYUSERPWD, 'username:password');
    curl_setopt($curl, CURLOPT_PROXYPORT, 'portno');
    $result = curl_exec($curl);
    curl_close($curl);
    return $result;
}
?>