Question

$sql = "SELECT * FROM users ORDER BY RAND() LIMIT 100";
$xml = new SimpleXMLElement('<xml/>');
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    while ($row = $result->fetch_array()) {
        $users = $xml->addChild('users');
        $users->addChild('Id',$row['sno']);
        $users->addChild('lat',$row['lat']);
        $users->addChild('long',$row['lng']);
        $users->addChild('address',$row['address']);
        }
$conn->close();   
//Create the XML file
    $fp = fopen("users100r.xml","wb");
//Write the XML nodes
    fwrite($fp,$xml->asXML());
    fclose($fp);
    echo $xml->saveXML();
?>

我想使用上面的代码创建一个XML文件，但问题是它没有生成我实际需要的必需结构。

我需要一个具有以下结构的XML文件。我该如何生成它？

<users>
<point id="1" lat="24.24707031" lng="68.16540527" address="Pakistan"/>
<point id="2" lat="34.24707031" lng="63.16540527" address="Lahore"/>
<point id="3" lat="28.24707031" lng="55.16540527" address="Karachi"/>
</users>

Answer 1

如果您希望users作为根元素，则必须以这种方式创建SimpleXMLElement。在循环中添加point元素并设置属性。

$rows = [
  ['id' => 1],
  ['id' => 2]
];

$users = new SimpleXMLElement('<users/>');
foreach ($rows as $row) {
  $point = $users->addChild('point');
  $point['id'] = $row['id'];
}

echo $users->asXml();

输出：

<?xml version="1.0"?>
<users><point id="1"/><point id="2"/></users>

如果您需要/需要更多控制，则必须切换到DOM。

$dom = new DOMDocument();
$users = $dom->appendChild($dom->createElement('users'));
foreach ($rows as $row) {
  $point = $users->appendChild($dom->createElement('point'));
  $point->setAttribute('id', $row['id']);
}

echo $dom->saveXml();

或者XMLWriter，这对大文件更好。

$xml = new XMLWriter();
$xml->openURI('php://output');
$xml->startDocument();
$xml->startElement('users');
foreach ($rows as $row) {
  $xml->startElement('point');
  $xml->writeAttribute('id', $row['id']);
  $xml->endElement();
}
$xml->endElement();
$xml->endDocument();

如何使用Php和Mysql创建具有特定结构的xml文件？

1 个答案: