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php 如何发送xml报文

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Release: 2016-06-23 14:04:39
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   POST <URL> HTTP/1.0  USER-Agent: <Client Application Name>  Content-Type: application/x-fox  Content-Length: nnnn    <?xml version=”1.0”?>  <?FOX FOXHEADER=”100” VERSION=”100” SECURITY=”NONE” LANG=CHS?>  <FOX>  ...  </FOX>
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现在要发送如上格式的xml数据,试过curl但不成功,大家帮帮忙


回复讨论(解决方案)

发到哪?你的代码是如何写的?

CURL是可以的,但是参数设置比较麻烦。这种情况有一些现成的类库实现,提供一个简单的示例

<?phpif( $_SERVER['REQUEST_METHOD'] === 'POST' ){    // 接收    $content = file_get_contents('php://input');    $xml = simplexml_load_string($content);    echo "来自XML接收方的响应\n";    print_r( get_object_vars($xml) );    exit;}// 发送行为$xml = <<<xml<?xml version="1.0"?><FOX><hello>world</hello></FOX>xml;$setting = array(    'http' => array(    'method' => 'POST',    'user_agent' => '<Client Application Name>',    'header' => "Content-type: application/x-www-form-urlencoded",    'content' => $xml    ));$context = stream_context_create($setting);$url = 'http://localhost/'. $_SERVER['REQUEST_URI'];$response = file_get_contents($url, null, $context);echo $response;
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需要说明你的HTTP请求示例中,请求内容实体与报头需要相隔两组\r\n。另外你的XML文件格式不正确,而且标签属性里还包含全角符号

CURL是可以的,但是参数设置比较麻烦。这种情况有一些现成的类库实现,提供一个简单的示例

PHP code?1234567891011121314151617181920212223242526272829303132
有什么现成的类库?

有用过fsockopen的吗?这个好像可以设置http报头

清明放假了?没人了?

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