Question

我想从xml获取架构元素列表。例如

<xs:schema id="SPPOnlineXML" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
 <xs:element name="SPPOnlineXML" msdata:IsDataSet="true" msdata:MainDataTable="lev_settings" msdata:UseCurrentLocale="true">
  <xs:complexType>
    <xs:choice minOccurs="0" maxOccurs="unbounded">
      <xs:element name="lev_settings">
        <xs:complexType>
          <xs:sequence>
            <xs:element name="LEVID" type="xs:int" minOccurs="0" />
            <xs:element name="ATTHEADID" type="xs:int" minOccurs="0" />
            <xs:element name="MINDATETYPE" type="xs:unsignedByte" minOccurs="0" />
            <xs:element name="MINMONTH" type="xs:unsignedByte" minOccurs="0" />
          </xs:sequence>
        </xs:complexType>
      </xs:element>
    </xs:choice>
  </xs:complexType>
 </xs:element>
</xs:schema>

从上面的模式我想创建一个元素数组＆＃34; LEVID＆＃34;，＆＃34; ATTHEAD＆＃34;，＆＃34; MINDATETYPE＆＃34;，＆＃34; MINMONTH＆＃34;

Answer 1

要将所有 xs:element name-attribute values 作为数组获取，我建议您选择SimpleXML并使用xpath。一旦你在字符串（或文件）中获得了XML，如果你知道xpath表达式，它就非常简单：

# Obtain all element names incl. complexTypes:

//xs:element/@name

# Obtain all element names excl. complexTypes and those
#  which contain anything incl. comments, text etc.:

//xs:element[not(node())]/@name

并且每个PHP代码例如：

$xml = simplexml_load_string($string);

echo "Obtain all element names incl. complexTypes:\n";

$elementNames = array_map('strval', $xml->xpath('//xs:element/@name'));
print_r($elementNames);

echo "\nObtain all element names excl. complexTypes and those
  which contain anything incl. comments, text etc.:\n";

$elementNames = array_map('strval', $xml->xpath('//xs:element[not(node())]/@name'));
print_r($elementNames);

See the online Demo.

Answer 2

代码：

$xml = new DOMDocument();

$xml->load('test.xml');
$xpath = new DOMXPath($xml);
$result = $xpath->query('//xs:sequence/xs:element');
$res_array = array();
foreach($result as $element_node) {
        $res_array[] = $element_node->getAttribute('name');
}
var_dump($res_array);

结果：

array(4) {
  [0]=>
  string(5) "LEVID"
  [1]=>
  string(9) "ATTHEADID"
  [2]=>
  string(11) "MINDATETYPE"
  [3]=>
  string(8) "MINMONTH"
}

Answer 3

我需要在html表上获取XSD文件的所有xs：元素，完整代码：

<table>
                <?php
                    // Load from xsd file
                    $url = "myFile.xsd";
                    $xml = file_get_contents($url, FILE_USE_INCLUDE_PATH);
                    $xml = simplexml_load_string($xml);


                    $elementNames = array_map('strval', $xml->xpath('//xs:element/@name'));
                    foreach ($elementNames as $value) {
                        echo "<tr><td>$value</td></tr>";
                    }

                    $elementNames = array_map('strval', $xml->xpath('//xs:element[not(node())]/@name'));
                    foreach ($elementNames as $value) {
                        echo "<tr><td>--->$value</td></tr>";
                    }
                ?>
</table>

如果您正在寻找xs：属性，您只需要将所有代码替换为＆＃34; xs：element＆＃34; by＆＃34; xs：attribute＆＃34;

如何从PHP中的XMLSchema XML文件中读取元素名称数组？

3 个答案: