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如何將此java代碼轉換爲php。將java rest請求轉換爲php代碼
server = new URL(url);
HttpURLConnection urlConnection = (HttpURLConnection) server.openConnection() ;
urlConnection.setRequestMethod(method);
urlConnection.setDoInput(true);
urlConnection.setDoOutput(true);
urlConnection.setRequestProperty("Content-Type",mimeType);
if(requestParameter != null)
{
CyberoamLogger.appLog.debug(MODULE+":"+METHOD+"Request parameter: "+requestParameter.toString());
out = new BufferedWriter(new OutputStreamWriter(urlConnection.getOutputStream()));
out.write(requestParameter.toString());
out.flush();
out.close() ;
}
urlConnection.connect();
strbuf = new StringBuffer();
is= urlConnection.getInputStream() ;
byte[] buffer = new byte[1024 * 4];
int n = 0;
while (-1 != (n = is.read(buffer)))
strbuf.append(new String(buffer, 0, n));
is.close();
strResponse=strbuf.toString();
CyberoamLogger.appLog.debug(MODULE+METHOD+ " WS Responce in String : "+strResponse);
urlConnection.disconnect();
,當我嘗試下面的PHP代碼我得到的,因爲請求實體是()不是由所請求的資源請求方式所支持的格式,服務器拒絕了這個請求。錯誤
$curl = curl_init();
curl_setopt($curl, CURLOPT_POSTFIELDS, $data);
curl_setopt($curl, CURLOPT_URL, $url);
echo curl_exec($curl);
mime類型是包含Java變量「應用程序/ JSON」。在PHP中,當我把CURLOPT_HTTPHEADER數組(「Content-Type:application/json」); –
它給我錯誤客戶端發送的請求在語法上不正確()。 –
@SiddharthNagar您的json數據格式可能不正確。檢查如何從我更新的答案中使用正確的格式化json。 –