Selenium/java-xpath for xlink:href attribute

Question

I tried to locate use element with attribute value '#aaa' for below snippet

<svg>
    <use xlink:href='#aaa'></use>
</svg>

I was able to locate using css selector use[* |href='#aaa'] . But I require it in xpath. I tried below xpath but it is not working

(//*[name()='use'])[ * ='#aaa' or @href='#aaa']

Note :I need to use the value '#aaa' to differentiate from other elements.

Could anyone please help to construct this xpath.

Xenophanes

Answer 1

Your predicate [ * ='#aaa' or @href='#aaa'] returns false as there is no attribute as href (but xlink:href), and use has no child nodes with text content '#aaa'.

Note that wild-card for any attribute is @* while just * is for any node (except attribute). Try below XPath

"//*[name()='use' and @*='#aaa']"

Cralg

FYI: This works for selenium in c# as well – Sunbeam

Answer 2

For XPath:

//*[name()='use'][contains(@*, '#aaa')]

Or variant provided by Andersson

Approaches with [@xlink:href='#aaa'] don't work on practice

Eulalia

Answer 3

as none of the suggestions really worked for me, I came up with following (tested in chrome devtools):

//*[name()='use' and @*[contains(.,'#aaa)]]

comments:

• name()='use'
  • because a normal tag search (i.e //use) did not work
• @*[contains(.,'#aaa')]
  • because contains with a wildcard (i.e. contains(@*,'#aaa')) didn't work

Reannareap

Answer 4

lookout, if someone needs contains..

"//*[name()='use'][contains(@*='#aaa')]"

Teem

Answer 5

This is an SVG tag and it works a little differently

Please try below xpath:

//[local-name()='svg']/yourxpath

Alamo

Could you please let me know whether it works or not – Alamo

It does not work. – Warman

Answer 6

Use the xpath

//use[@xlink:href='#aaa']

or

//use[contains(@xlink:href,'#aaa')]

Dunleavy

both suggestions end up invalid xpath expression – Lonesome

Answer 7

As the element is within svg namespace you can use the following xpath:

//*[name()="svg"]/use[contains(@xlink:href,'#aaa')]

Herzog

Answer 8

Try this who wants contains

//@*[name()='xlink:href' and contains(., '#aaa')]

Olson