PHP框架调用Java后端,参数传递不过去的问题解决
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2022-04-09 18:13:13
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本篇文章给大家分享的内容是PHP框架调用Java后端,参数传递不过去的问题解决,有着一定的参考价值,有需要的朋友可以参考一下
public function request($requestURL,$params='',$method ='GET',$contentType='',$user=''){ $timeout = 30; $ch = null; if ('POST' === strtoupper($method)) { $ch = curl_init($requestURL); curl_setopt($ch, CURLOPT_POST, 1); curl_setopt($ch, CURLOPT_FRESH_CONNECT, 1); curl_setopt($ch, CURLOPT_FORBID_REUSE, 1); if (is_string($params)) { curl_setopt($ch, CURLOPT_POSTFIELDS, $params); } else { curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($params)); } } else if('GET' === strtoupper($method)) { if(is_string($params)) { $real_url = $requestURL. (strpos($requestURL, '?') === false ? '?' : ''). $params; } else { $real_url = $requestURL. (strpos($requestURL, '?') === false ? '?' : ''). http_build_query($params); } $ch = curl_init($real_url); } else { $args = func_get_args(); return false; } if ($contentType) { curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type:'.$contentType)); } if ($user) { curl_setopt($ch, CURLOPT_USERPWD, $user); } curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt($ch, CURLOPT_TIMEOUT, $timeout); $ret = curl_exec($ch); $info = curl_getinfo($ch); $contents = array( 'httpInfo' => array( 'send' => $params, 'url' => $requestURL, 'ret' => $ret, 'http' => $info, ) ); curl_close($ch); return $ret; }
System.out.println("Content Type: " + request.getContentType());
此方法之前传递的Content Type为text/html
后面把传值去掉,为空,传递过去的默认值为application/x-www-form-urlencoded
就可以了。
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