Extracting the value of the attribute version in a geoserver SLD file with xmllint and xpath

Question

i need to extract the version of a SLD file for GeoServer, which is an XML-based markup language. The version is an attribute of the element StyledLayerDescriptor.

Here is the xml file :

$ cat my_geoserver_sld_file.sld
<?xml version="1.0" encoding="ISO-8859-1"?>
  <StyledLayerDescriptor version="1.0.0"
    xsi:schemaLocation="http://www.opengis.net/sld http://schemas.opengis.net/sld/1.0.0/StyledLayerDescriptor.xsd"
    xmlns="http://www.opengis.net/sld" xmlns:ogc="http://www.opengis.net/ogc"
    xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <NamedLayer>
      <Name>230_sld_shp_line__230_test_sld_shp_line</Name>
        <UserStyle>
        <Title>A green line style</Title>
        <FeatureTypeStyle>
          <Rule>
             <Title>green line</Title>
             <LineSymbolizer>
               <Stroke>
                 <CssParameter name="stroke">#00ff00</CssParameter>
               </Stroke>
             </LineSymbolizer>
          </Rule>
        </FeatureTypeStyle>
      </UserStyle>
    </NamedLayer>
  </StyledLayerDescriptor>

I would like to set : version="1.0.0"

first the file was opened with de "xmllint --shell" command, in order to use xpath:

$ xmllint --shell my_geoserver_sld_file.sld
/ > xpath *
Object is a Node Set :
Set contains 1 nodes:
1  ELEMENT StyledLayerDescriptor
    default namespace href=http://www.opengis.net/sld
    namespace ogc href=http://www.opengis.net/ogc
    namespace xlink href=http://www.w3.org/1999/xlink
    namespace xsi href=http://www.w3.org/2001/XMLSchema-instanc...
    ATTRIBUTE version
      TEXT
        content=1.0.0
    ATTRIBUTE schemaLocation
      TEXT
        content=http://www.opengis.net/sld http://schema...

it should be simple to extract the version, but it fails...

/ > cat //StyledLayerDescriptor/version/text()
/ >

How can i set version in a bash variable ?

Daniel Haley · Accepted Answer · 2018-05-08T16:36:05.870

1

Your XPath doesn't work because of the default namespace (http://www.opengis.net/sld) in your XML.

See this answer for some options on handling default namespaces in xmllint.

Additionally, since the attribute you're trying to select is on the root element, just use /* in your xpath...

xmllint --xpath "/*/@version" my_geoserver_sld_file.sld

This will return version="1.0.0". If you just want the value 1.0.0, use string()...

xmllint --xpath "string(/*/@version)" my_geoserver_sld_file.sld

edited May 08 '18 at 16:36

answered May 08 '18 at 16:30

Daniel Haley

51,389
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Thanks ! Note that string() is explained [here](https://stackoverflow.com/questions/11611385/how-can-i-get-the-value-from-an-attribute-using-xmllint-and-xpath/24051117) – MaxiReglisse May 09 '18 at 14:58

score 0 · Answer 2 · answered May 14 '18 at 09:51

0

as suggested by Daniel Halley, it is possible to use the local-name() to match the element name StyledLayerDescriptor:

xmllint --xpath "string(/*[local-name()='StyledLayerDescriptor']/@version)" my_geoserver_sld_file.sld

answered May 14 '18 at 09:51

MaxiReglisse

3,093
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Extracting the value of the attribute version in a geoserver SLD file with xmllint and xpath

2 Answers2