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PHP curl不使用异常字段发布_Php_Curl - Fatal编程技术网

PHP curl不使用异常字段发布

PHP curl不使用异常字段发布,php,curl,Php,Curl,我正在尝试使用curl来发布到这个站点,但我甚至无法打开页面。。字段不像有名称和值那样直接,表单有一个onsubmit。每次重新加载页面时,表单的地址和操作似乎也会改变,因此地址中“e”后面的数字每次增加1 $url = "http://www.canadapost.ca/cpotools/apps/track/personal/findByRefNumber?execution=e14s1" $parms = array( 'tapByTrackSearch:refSe

我正在尝试使用curl来发布到这个站点,但我甚至无法打开页面。。字段不像有名称和值那样直接,表单有一个onsubmit。每次重新加载页面时,表单的地址和操作似乎也会改变,因此地址中“e”后面的数字每次增加1

$url = "http://www.canadapost.ca/cpotools/apps/track/personal/findByRefNumber?execution=e14s1"
$parms = array(
            'tapByTrackSearch:refSearch:refNumbers' => $refnumber,
            'tapByTrackSearch:refSearch:ref_submit_button' => '',
            'tapByTrackSearch:refSearch_SUBMIT' => 1,
            'javax.faces.ViewState' => 'v1Dxx9hSieZwMVPzZZHAPSkIfrSZM1jbO4mNTTN2U0GpxoAFjzFhwFOL94NDwf6uXDCuuM24L5YMuqNSqP+dRYKpGsqhkV+2Tm+y5O2kdGF/NCNJRzUyCzDzaoELlvqQ/L5LRDJEnbwOXh1YOjylrA==',
                ); 
我稍后循环$parms数组,并作为$new_字段字符串传递

    $res    =   curl_init();
$config =   
array(
      CURLOPT_URL   =>  $this->url,
      CURLOPT_POST =>   TRUE,
      CURLOPT_POSTFIELDS => $new_fields,
      CURLOPT_RETURNTRANSFER => TRUE,
      CURLOPT_FOLLOWLOCATION => TRUE
    );

curl_setopt_array($res,$config);
$result = curl_exec($res);
curl_close($res);
echo $result;
新字段的输出:

tapByTrackSearch:refSearch:refNumbers=#12484&tapByTrackSearch:refSearch:ref_submit_button=&tapByTrackSearch:refSearch_SUBMIT=1&javax.faces.ViewState=v1Dxx9hSieZwMVPzZZHAPSkIfrSZM1jbO4mNTTN2U0GpxoAFjzFhwFOL94NDwf6uXDCuuM24L5YMuqNSqP+dRYKpGsqhkV+2Tm+y5O2kdGF/NCNJRzUyCzDzaoELlvqQ/L5LRDJEnbwOXh1YOjylrA==&

我的代码在其他网站上运行良好,但出于某种原因,我要么输入的字段及其值错误,要么输入的是url

var\u dump
$new\u fields
variableMy guess是
execution
参数是某种会话ID,它必须与服务器创建表单时发送的值相匹配。由于会话已结束,以后尝试重新使用它将不起作用。var\u dump:bool(false)是否有方法从客户端启动新会话?@joe将
$new\u字段的输出添加到问题中