php curl怎么用？

我用AJAX POST数据过去可以、、用curl怎么不行？
$ch = curl_init($url); 
curl_setopt($ch, CURLOPT_HEADER,0);
curl_setopt($ch,CURLOPT_POST,1); 
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); 
curl_setopt($ch,CURLOPT_POSTFIELDS,$curlPost); 
$data=curl_exec($ch); 
curl_close($ch); 

解决方案 »

  1.   


    $url="http://www.ems.com.cn/tools/postageenresult.jsp"; 
    $post_data=array(
        'reqCode'=>'Postageen',
        'goodsType'=>'goods',
        'addressto'=>'阿根廷',
        'weight'=>2);
    $data= http_build_query($post_data);
    $ch=curl_init($url); 
    curl_setopt($ch,CURLOPT_POST,1); 
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); 
    curl_setopt($ch,CURLOPT_POSTFIELDS,$data); 
    $result=curl_exec($ch); 
    curl_close($ch); 
    print_r($result); 
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>无标题文档</title>
    </head>
    <script language="javascript">
    function makeRequest() {
    http_request = false; if (window.XMLHttpRequest) {
    http_request = new XMLHttpRequest();

    if (http_request.overrideMimeType) {
    http_request.overrideMimeType('text/xml');
    }
    } else if (window.ActiveXObject) { // IE
       
    try {http_request = new ActiveXObject("Msxml2.XMLHTTP");} catch (e) {}

    } if (!http_request) {
    alert('不能创建 XMLHTTP 实例');
    return false;
    }
    http_request.onreadystatechange = showfeeContents;
        var urls="http://www.ems.com.cn/tools/postageenresult.jsp";
    http_request.open('POST', urls, true);
    http_request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); var sendurl="";
    var address="阿根廷";
    var weights=2;
            sendurl="reqCode=Postageen&goodsType=goods&addressto="+address+"&weight="+weights;
    http_request.send(sendurl);

    }
    function showfeeContents() {
    if (http_request.readyState == 4) { if (http_request.status == 200) {
        var test=http_request.responseText;
    alert(test);
    }
    else
    {
    alert('请求出错121.');
    }

    }
    }
    </script>
    <body>
    <span style=" cursor:pointer;" onclick="makeRequest()" >POST</span>
    <div id="price"></div>
    </body>
    </html>Php的不行，html可以、、