PHP框架调用Java后端，参数传递不过去的问题解决

 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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