使用php代理的跨域Ajax GET请求

在过去的几天里，我一直把头发拉过来。 我试图从公共API获得跨域ajax GET请求，但我根本无法让它工作。

PHP：我正在使用Ben Alman的ba-simple-proxy（ https://github.com/cowboy/php-simple-proxy ）

 'ERROR' ); } else if ( !preg_match( $valid_url_regex, $url ) ) { // Passed url doesn't match $valid_url_regex. $contents = 'ERROR: invalid url'; $status = array( 'http_code' => 'ERROR' ); } else { $ch = curl_init( $url ); if ( strtolower($_SERVER['REQUEST_METHOD']) == 'post' ) { curl_setopt( $ch, CURLOPT_POST, true ); curl_setopt( $ch, CURLOPT_POSTFIELDS, $_POST ); } if ( $_GET['send_cookies'] ) { $cookie = array(); foreach ( $_COOKIE as $key => $value ) { $cookie[] = $key . '=' . $value; } if ( $_GET['send_session'] ) { $cookie[] = SID; } $cookie = implode( '; ', $cookie ); curl_setopt( $ch, CURLOPT_COOKIE, $cookie ); } curl_setopt( $ch, CURLOPT_FOLLOWLOCATION, true ); curl_setopt( $ch, CURLOPT_HEADER, true ); curl_setopt( $ch, CURLOPT_RETURNTRANSFER, true ); curl_setopt( $ch, CURLOPT_USERAGENT, $_GET['user_agent'] ? $_GET['user_agent'] : $_SERVER['HTTP_USER_AGENT'] ); list( $header, $contents ) = preg_split( '/([\r\n][\r\n])\\1/', curl_exec( $ch ), 2 ); $status = curl_getinfo( $ch ); curl_close( $ch ); } // Split header text into an array. $header_text = preg_split( '/[\r\n]+/', $header ); if ( $_GET['mode'] == 'native' ) { if ( !$enable_native ) { $contents = 'ERROR: invalid mode'; $status = array( 'http_code' => 'ERROR' ); } // Propagate headers to response. foreach ( $header_text as $header ) { if ( preg_match( '/^(?:Content-Type|Content-Language|Set-Cookie):/i', $header ) ) { header( $header ); } } print $contents; } else { // $data will be serialized into JSON data. $data = array(); // Propagate all HTTP headers into the JSON data object. if ( $_GET['full_headers'] ) { $data['headers'] = array(); foreach ( $header_text as $header ) { preg_match( '/^(.+?):\s+(.*)$/', $header, $matches ); if ( $matches ) { $data['headers'][ $matches[1] ] = $matches[2]; } } } // Propagate all cURL request / response info to the JSON data object. if ( $_GET['full_status'] ) { $data['status'] = $status; } else { $data['status'] = array(); $data['status']['http_code'] = $status['http_code']; } // Set the JSON data object contents, decoding it from JSON if possible. $decoded_json = json_decode( $contents ); $data['contents'] = $decoded_json ? $decoded_json : $contents; // Generate appropriate content-type header. $is_xhr = strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'; header( 'Content-type: application/' . ( $is_xhr ? 'json' : 'x-javascript' ) ); // Get JSONP callback. $jsonp_callback = $enable_jsonp && isset($_GET['callback']) ? $_GET['callback'] : null; // Generate JSON/JSONP string $json = json_encode( $data ); print $jsonp_callback ? "$jsonp_callback($json)" : $json; } ?> 

我试图获取信息的API是aviationweather API。 端点url：

 https://www.aviationweather.gov/adds/dataserver_current/httpparam?datasource=metars&requestType=retrieve&format=xml&mostRecentForEachStation=constraint&hoursBeforeNow=1.25 

jQuery的：

 function getMetarToTable(){ var proxy = 'simple-proxy.php', url=proxy + '?' +'https://www.aviationweather.gov/adds/dataserver_current/httpparam?datasource=metars&requestType=retrieve&format=xml&mostRecentForEachStation=constraint&hoursBeforeNow=1.25'; var stationt="KJFK"; $.ajax({ type: "GET", url: url, crossDomain: true, data: {'stationString': stationt}, datatype: "xml", error: function(jqXHR, textStatus, errorThrown) { console.log('Error: ' + errorThrown); }, success: function(xml) { $("#metarweatherpage").html(''); $(xml).find('METAR').each(function(){ var sRawtext = $(this).find('raw_text').text(); var sStationid = $(this).find('station_id').text(); $("#metarweatherpage").append('
'+sStationid+'
'+sRawtext+'
'); }); }, }); }; 

我真的很感激我能得到的任何帮助。

Answer 1

从PHP：

 $url = $_GET['url']; 

它试图从名为url的查询字符串参数中读取URL。

来自JavaScript：

 url=proxy + '?' +'https://www.aviationweather.gov/adds/dataserver_current/httpparam?datasource=metars&requestType=retrieve&format=xml&mostRecentForEachStation=constraint&hoursBeforeNow=1.25'; 

您正在传递数据raw，而不将其编码为查询字符串。

要做到这一点：

 var url = 'https://www.aviationweather.gov/adds/dataserver_current/httpparam?datasource=metars&requestType=retrieve&format=xml&mostRecentForEachStation=constraint&hoursBeforeNow=1.25' var query = encodeURIComponent("url") + "=" + encodeURIComponent(url); var proxy_url = proxy + "?" + query; 

---

也就是说，您已经在使用jQuery，因此您不应该首先手动构建URL：

 var url = 'https://www.aviationweather.gov/adds/dataserver_current/httpparam?datasource=metars&requestType=retrieve&format=xml&mostRecentForEachStation=constraint&hoursBeforeNow=1.25' $.ajax({ type: "GET", url: proxy, data: {'url': url}, // etc