How to use like in XPath?

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Solution 1

You can use all of the XPath (1.0) string functions. If you have XPath 2.0 available, then you can even use RegEx.

contains()

starts-with()

substring()

substring-before()

substring-after()

concat()

translate()

string-length()

There is no **ends-with() in XPath 1.0, but it can easily be expressed with this XPath 1.0 expression**:

substring($s, string-length($s) - string-length($t) +1) = $t

is true() exactly when the string $s ends with the string $t.

Solution 2

You can use start-with function and not function. Reference:

http://www.w3schools.com/xpath/xpath_functions.asp

xmlTempResultSearch = xmlResidentListDisplay.selectNodes("//PeopleList/Row[not(starts-with(@LastName,'"+ txtSearch.value +"'))]");

Solution 3

you can use contains() function of XPath:

xmlTempResultSearch = xmlResidentListDisplay.selectNodes("//PeopleList/Row[not(contains(@LastName,'"+txtSearch.value+"'))]");
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edsamiracle
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edsamiracle

Updated on June 19, 2020

Comments

  • edsamiracle
    edsamiracle over 3 years

    I have a page that searches with filters. I have this code for example,

    xmlTempResultSearch = xmlResidentListDisplay.selectNodes("//PeopleList/Row[@LastName != '"+txtSearch.value+"']");
    xmlTempResultSearch.removeAll();
    

    This selects the data that is not equal to the LastName inputted on the txtSearch textbox and then removes them from the result set so that its filtered to equal the last name on the txtSearch textbox.

    My problem with this code is that it should be equal (=) to the txtSearch.value, what I want is that I want the result set LIKE the txtSearch.value. What happens on my page is that when I type 'santos' on the txtSearch textbox, its result set is all those last names with 'santos'. But when I type 'sant', nothing appears. I want the same result set with 'santos' because it all contains 'sant'

  • edsamiracle
    edsamiracle about 11 years
    I can't get this to work. I was searching for 'bro' since the last name I want to get is 'Broerman'. I got an error that says: Error: Expected token ']' found 'NAME'. //PeopleList/Row[not-->contains<--(@LastName, 'bro')]
  • Khue Vu
    Khue Vu about 11 years
    Oh, I am sorry, this should be not(contains(...))