Question

我迫切需要知道如何阅读以下XML文件：

<uclassify xmlns="http://api.uclassify.com/1/ResponseSchema" version="1.00">
<status success="true" statusCode="2000"/>
  <readCalls>
    <classify id="cls1">
      <classification>
        <class className="negative" p="0.741735"/>
        <class className="positive" p="0.258265"/>
      </classification>
    </classify>
  </readCalls>
</uclassify>

我需要知道以下内容：

$status_code = ... (should be 2000)
$negative = ...  (the value of p, should be 0.741735)
$positive = ...  (the value of p, should be 0.258265)

祝你好运

安德烈

Answer 1

$xml= simplexml_load_file('temp.xml');

foreach($xml->status->attributes() as $name => $value ){
    echo $name.' '.$value.'<br>';
    }
foreach($xml->readCalls->classify->classification->children() as $node ){
    foreach($node->attributes() as $name => $value)
     echo $name.' '.$value.'<br>';
    }

O / P：

成功是真的 statusCode 2000 className否定 p 0.741735 className为正 p 0.258265

如果您想存储它们：

$status_code = (string)$xml->status->attributes()->statusCode;
foreach($xml->readCalls->classify->classification->class as $node ){
    ${(string) $node->attributes()->className} = (string) $node->attributes()->p;
    }
echo $status_code.' '.$positive.' '.$negative;

O / P：

2000 0.258265 0.741735

如何从PHP中的复杂XML文件中提取各种值？

1 个答案: