Hi Paul,
getData() takes:
string $xPathQuery
xpath to the node - resolves to *one* xpath.
So the error message is saying that "//directory[name=help]/logo" matches
more than one node (or no node). So it doesn't know which node to return
you the data of. Try:
$aMatches = $xpath->match("//directory[name='help']/logo");
print_r($aMatches);
$logo= $xpath->getData($aMatches[0]);
echo $logo;
(Note the ' and the call to match before the call to getData).
You might get away with:
$logo= $xpath->getData("//directory[name='help']/logo");
echo $logo;
Cheers
Nigel
----- Original Message -----
From: "Paul Arnold" <Pa...@pi...>
To: <nig...@us...>
Sent: Thursday, April 24, 2003 4:31 PM
Subject: php.XPath class question
Hi Nigel,
I wonder if you can help me at all.
If I try to carry out
$logo= $xpath->getData("//directory[name=help]/logo");
echo $logo;
or
$logo= $xpath->getData("//directory[@name=help]/logo");
echo $logo;
on the piece of xml below I get the following error:
XPath error in XPath.class.php:5061 The supplied xPath '' does not
*uniquely* describe a node in the xml document.
Is this correct? Is it a bug?
Can you tell me how I select the //directory[@name=help]/logo node?
<root>
<directory name="company">
<logo>images/logo.gif</logo>
</directory>
<directory name="help">
<logo>images/logohelp.gif</logo>
</directory>
</root>
Kind regards,
Paul Arnold
Software Engineer
Pivetal Ltd.
Tel: 023 8021 5302
Email: pa...@pi...
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