Remove stylesheet from XML with PHP

Question

I'm tring to show some XML for Italian Electronic invoices received, using my custom stylesheet.xsl

All is ok when XML received start with:

<?xml version="1.0" encoding="UTF-8"?>
<p:FatturaElettronica xmlns:p="http://ivaservizi.agenziaentrate.gov.it/docs/xsd/fatture/v1.2" versione="FPR12">

but I've received some XML starting with:

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="fatturapa_v1.2.xsl"?>
<p:FatturaElettronica xmlns:p="http://ivaservizi.agenziaentrate.gov.it/docs/xsd/fatture/v1.2" versione="FPR12">

in this case i get browser error when I try to open file because on my webapp i have not the fatturapa_v1.2.xml saved:

Error loading style sheet: XSLT style sheet interpretation failed.

Is there a way to strip out from this XML this line only, using PHP? Thanks

<?xml-stylesheet type="text/xsl" href="fatturapa_v1.2.xsl"?>

You can try using SimpleXML. Not sure how it deals with stylesheets though. — M. Eriksson
– M. Eriksson, Commented Jan 24, 2022 at 15:14

ThW · Accepted Answer · 2022-01-24 18:09:11Z

1

Everything in DOM is a node. In this case this is a processing instruction. You can use Xpath to find it and then remove it from its parent node:

$xml = <<<'XML'
<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="fatturapa_v1.2.xsl"?>
<p:FatturaElettronica xmlns:p="http://ivaservizi.agenziaentrate.gov.it/docs/xsd/fatture/v1.2" versione="FPR12"/>
XML;

$document = new DOMDocument();
$document->loadXML($xml);
$xpath = new DOMXpath($document);

foreach ($xpath->evaluate('/processing-instruction()[name() = "xml-stylesheet"]') as $pi) {
    // var_dump($pi);
    $pi->parentNode->removeChild($pi);
}

echo $document->saveXML();

answered Jan 24, 2022 at 18:09

ThW

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