Question

我正在尝试从http://ratings.food.gov.uk/OpenDataFiles/FHRS523en-GB.xml访问xml数据，但似乎无法正确检索它。但我不认为它与数据结构有任何关系。如何检索和显示数据？

<?php
$fhrsid = 'FHRSID';

$LocalAuthorityBusinessID = 'LocalAuthorityBusinessID';

$xml = simplexml_load_file("http://ratings.food.gov.uk/OpenDataFiles/FHRS523en-GB.xml");
echo "<h2>".$xml->getName()."</h2><br />";

foreach($xml->children() as $data)

{
    echo "FHRSID: ".$data->$fhrsid."<br />";
    echo "LocalAuthorityBusinessID : ".$data->$LocalAuthorityBusinessID." <br />";
    echo "BusinessName : ".$data->BusinessName." <br />";
    echo "BusinessType : ".$data->BusinessType." <br />";
    echo "BusinessTypeID : ".$data->BusinessTypeID." <br />";
    echo "AddressLine1 : ".$data->AddressLine1." <br />";
    echo "AddressLine2 : ".$data->AddressLine2." <br />";
    echo "<hr/>";
}
?>

Answer 1

执行：

print_r($xml);

输出：

SimpleXMLElement Object
(
    [Header] => SimpleXMLElement Object
        (
            [ExtractDate] => 2015-04-25
            [ItemCount] => 1881
            [ReturnCode] => Success
        )

    [EstablishmentCollection] => SimpleXMLElement Object
        (
            [EstablishmentDetail] => Array
                (
                    //...

您将看到xml的结构，您将看到必须将foreach标题更改为：

foreach($xml->EstablishmentCollection->EstablishmentDetail as $data)

修改

同样如comments中的@hakre所述，您应该使用asXML()：

echo $xml->asXML();

查看完整的整个xml。

使用PHP显示XML

1 个答案: