根据我已有的原始数据,我希望发出一个HTTP post请求。我花了一段时间寻找解决方案,做了几次尝试,我正在寻求一点帮助。我要做的PHP代码如下所示:
<?
$url="http://localhost:3000";
$postdata="<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<hi></hi>";
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $postdata);
$result = curl_exec($ch);
curl_close($ch);
echo($result);
?>我的尝试是这样的:
private String setXmlPostHeader(Document doc, PostMethod postMethod) throws java.io.IOException, java.io.UnsupportedEncodingException,
javax.xml.transform.TransformerException
{
ByteArrayOutputStream xmlBytes = new ByteArrayOutputStream();
XML.serialize( doc, xmlBytes );
final byte[] ba = xmlBytes.toByteArray();
String data = new String(ba, "utf-8");
InputStreamRequestEntity re = new InputStreamRequestEntity(new ByteArrayInputStream(ba));
postMethod.setRequestEntity(re);
postMethod.setRequestHeader("Content-type", MediaType.XML.toString() + "; charset=UTF-8");
return data;
}然后执行postMethod,但这只是一个不包含任何数据的post。有没有人看到我做错了什么?我想弄清楚如何改变这个方法,使它真正工作。谢谢!
-Ken
发布于 2010-08-05 09:25:37
java.net.URLConnection类不是工作得更好吗?
发布于 2010-08-05 09:58:02
看起来你不像是在打电话:
int result = httpclient.executeMethod(postMethod);
postMethod.releaseConnection();https://stackoverflow.com/questions/3411093
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