有没有更简洁的方法来实现 NHS 数据从 XML 到 JSON 的转换
我正在使用来自公共 NHS API 的位置和地址数据，它以 XML 格式输入，我已将其转换为 JSON 以在我的 web 应用程序中使用。
我的代码正在运行并产生所需的结果，但是我不禁对这段代码中嵌套的 foreach 循环的数量感到畏缩。
有没有更干净的方法来实现相同的结果，并且性能更高？
我曾尝试使用 simplexml_load_string 和许多变体，但它拒绝输出/解析包含 s:organisationSummary 数据的嵌套元素的内容。
这是 PHP 类
class XmlElement {
public $name;
public $attributes;
public $content;
public $children;
}
class GpService
{
protected $apiKey;
public function __construct()
{
$this->apiKey = env('NHS_API_KEY');
}
public function xmlToObject($xml) {
$parser = xml_parser_create();
xml_parser_set_option($parser, XML_OPTION_CASE_FOLDING, 0);
xml_parser_set_option($parser, XML_OPTION_SKIP_WHITE, 1);
xml_parse_into_struct($parser, $xml, $tags);
xml_parser_free($parser);
$elements = array(); // the currently filling [child] XmlElement array
$stack = array();
foreach ($tags as $tag) {
$index = count($elements);
if ($tag['type'] == "complete" || $tag['type'] == "open") {
$elements[$index] = new XmlElement;
$elements[$index]->name = $tag['tag'];
$elements[$index]->attributes = (isset($tag['attributes']))?$tag['attributes']:null;
$elements[$index]->content = (isset($tag['value']))?$tag['value']:null;
if ($tag['type'] == "open") { // push
$elements[$index]->children = array();
$stack[count($stack)] = &$elements;
$elements = &$elements[$index]->children;
}
}
if ($tag['type'] == "close") { // pop
$elements = &$stack[count($stack) - 1];
unset($stack[count($stack) - 1]);
}
}
return $elements[0]; // the single top-level element
}
}
}
慕尼黑56888552回答
-
繁花不似锦
您可以使用 DOM 和 XPath 表达式。所有数据都在{https://api.nhs.uk/data/services}organisationSummary节点内。下面的示例注册service命名空间的前缀https://api.nhs.uk/data/services（这就是s示例中的 解析为）。然后它{https://api.nhs.uk/data/services}organisationSummary使用//service:organisationSummary表达式获取任何节点。对于每个组织，它$item使用标量值的表达式构建一个变量。循环用于{https://api.nhs.uk/data/services}addressLine节点。// create the DOM document and load the XML $document = new DOMDocument();$document->loadXML($xml);// create an Xpath instance for the document - register namespace$xpath = new DOMXpath($document);$xpath->registerNamespace('service', 'https://api.nhs.uk/data/services');$json = [ "success" => true, // force object for data - it will be an array otherwise "data" => new stdClass()];// iterate over all organisationSummary nodesforeach ($xpath->evaluate('//service:organisationSummary') as $index => $organisation) { // fetch single values $item = [ "name" => $xpath->evaluate('string(service:name)', $organisation), "odscode" => $xpath->evaluate('string(service:odsCode)', $organisation), "postcode" => $xpath->evaluate('string(service:address/service:postcode)', $organisation), "telephone" => $xpath->evaluate('string(service:contact/service:telephone)', $organisation), "longitude" => $xpath->evaluate('string(service:geographicCoordinates/service:longitude)', $organisation), "latitude" => $xpath->evaluate('string(service:geographicCoordinates/service:latitude)', $organisation), "distance" => $xpath->evaluate('string(service:Distance)', $organisation) ]; // add the addressLine values foreach ($xpath->evaluate('service:address/service:addressLine', $organisation) as $lineIndex => $line) { $item['addressline'.$lineIndex] = $line->textContent; } // add the $item $json['data']->{$index} = $item;}echo json_encode($json, JSON_PRETTY_PRINT);输出：{ "success": true, "data": { "0": { "name": "Highfields Surgery (R Wadhwa)", "odscode": "C82116", "postcode": "LE2 0NN", "telephone": "01162543253", "longitude": "-1.11859357357025", "latitude": "52.6293983459473", "distance": "0.247038430918239", "addressline0": "25 Severn Street", "addressline1": "Leicester", "addressline2": "Leicestershire" }, "1": { "name": "Dr R Kapur & Partners", "odscode": "C82659", "postcode": "LE2 0TA", "telephone": "01162951258", "longitude": "-1.11907768249512", "latitude": "52.6310386657715", "distance": "0.30219913005026", "addressline0": "St Peters Health Centre", "addressline1": "Sparkenhoe Street", "addressline2": "Leicester", "addressline3": "Leicestershire" }, ... -
冉冉说
您需要使用适当的方法来遍历 XML 文档，而不是查看每个元素并检查其名称。您的两个最佳选择是 DOM 方法和 XPath。如果您使用过前端 JavaScript 代码（例如document.getElementsByTagName或document.getElementById），您就会熟悉前者。后者具有更陡峭的学习曲线，但功能要强大得多。（XSLT 也是转换 XML 的好选择，但我对它不是很熟悉。）不过，第一步是使用正确的 XML 库。您使用的 XML 函数级别非常低。我建议改用 PHP 的DomDocument扩展。让我们潜入吧！<?php$xml = file_get_contents("nhs.xml");// assume XML document is stored in a variable called $xml$dom = new DomDocument;$dom->loadXml($xml);// thankfully we can ignore namespaces!$summaries = $dom->getElementsByTagName("organisationSummary");// go through each <organisationSummary> elementforeach ($summaries as $os) { $entry_data = [ // note we're using $os and not $dom now, so we get child elements of this particular element "name" => $os->getElementsByTagName("name")[0]->textContent, "odscode" => $os->getElementsByTagName("odsCode")[0]->textContent, "addressline0" => $os->getElementsByTagName("addressLine")[0]->textContent, // provide a default empty string in case there's no further lines "addressline1" => $os->getElementsByTagName("addressLine")[1]->textContent ?? "", "addressline2" => $os->getElementsByTagName("addressLine")[2]->textContent ?? "", "addressline3" => $os->getElementsByTagName("addressLine")[3]->textContent ?? "", "postcode" => $os->getElementsByTagName("postcode")[0]->textContent, "telephone" => $os->getElementsByTagName("telephone")[0]->textContent, "longitude" => $os->getElementsByTagName("longitude")[0]->textContent, "latitude" => $os->getElementsByTagName("latitude")[0]->textContent, "distance" => $os->getElementsByTagName("Distance")[0]->textContent, ]; $json_data[] = $entry_data;}if (count($json_data)) { $output = ["success" => true, "data" => $json_data];} else { $output = ["success" => false];}echo json_encode($output, JSON_PRETTY_PRINT);
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