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Said's avatar
Said
Contributor
6 years ago
Solved

How to select nodes matching criteria

Hi all, Because large number of <c> elements in the response my checks take very long (hours). See simplified respresentation of response below. The actual response has much more elements. <a> <...
  • groovyguy's avatar
    groovyguy
    6 years ago

    Said: That's what I was afraid of. Unfortunately if you have to calculate each and every value that exists in the /b/c node where E exists, then you have to process every node individually and that is going to take time. I ran into similar problems because, as you stated, the arrays that the statements I provided would be completely and totally out of order and out of sync. 

     

    In my experience, there's no good way to do this other than parsing every //b/c node. 

groovyguy's avatar
groovyguy
Champion Level 1
6 years ago

 

Unfortunately, as far as I am aware, this is highly dependent on what you are doing and checking if this snippet returns true. 

exists(//b/c[$i]/e)

 

Can you give us an example of what you are doing in that scenario? If you need to check the value of D or F if E exists, you can play with an XPATH statement like below. This says, for the entire response, get me the value of //b/c/d where //b/c/e exists. It returns it as an array. 

 

log.info(holder["//b/c[exists(./e)]/d/text()"].toString());

This outputs [1, 4, 9, 15] based on the sample response you gave. 

  • Said's avatar
    Said
    Contributor
    6 years ago

    groovyguy, thanks alot for you fast response :smileyhappy:

    To answer you question: Can you give us an example of what you are doing in that scenario?

    I need to copy the values of d, e, f (and more, I left them out to keep it to the point) to an in memory database, to that I can use them to verify a calculation.

    Below statement gives me an array of all d when e exists.

    String[] d = respHolder["//b/c[exists(./e)]/d/text()"];

     I could use same type of statement to create and array for e and f. But what if e.g. f is not there in one of the nodes. The arrays are then not in sync. How could I deal with this?

    • groovyguy's avatar
      groovyguy
      Champion Level 1
      6 years ago

      Said: That's what I was afraid of. Unfortunately if you have to calculate each and every value that exists in the /b/c node where E exists, then you have to process every node individually and that is going to take time. I ran into similar problems because, as you stated, the arrays that the statements I provided would be completely and totally out of order and out of sync. 

       

      In my experience, there's no good way to do this other than parsing every //b/c node. 

      • Said's avatar
        Said
        Contributor
        6 years ago

        @groovyguy At least I know how to proceed with this now! The array is a great idea :smileyvery-happy:

        Thanks alot.