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Question

Question

Parameter in URL search string

Web Client
Updated June 8, 2017
asked on June 7, 2017 • Show version history

I have an application that has the capability to pop out to a specified URL.  So I've built a search string for Laserfiche and want to use the Laserfiche search string (which is a fully qualified URL) as the URL in my other application.

The problem is this:  how do I code a variable value in the external app and embed that value in the Laserfiche URL?  I would need a parameter in the Laserfiche search string but how do I code that parameter?

Example: 

https://lfweb.ourdomain.us/laserfiche/Browse.aspx?repo=OurRepository#?view=search&search=%7BLF:Basic%20~%3D%20%22311-15-123%22,%20option%3D%22DFNLT%22%7D%20%26%20%7BLF:Name%3D%22*%22,%20Type%3D%22DF%22%7D%20%26%20%7BLF:LOOKIN%3D%22%5CAssessor%5CPRC%22%7D

In this URL is the embedded search value "311-15-123", but how do I code a parameter in the URL that will "catch" the value being passed from the other app?  So basically what I need to do is replace the hardcoded string "311-15-123" with a parameter.  Is this possible?

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Replies

replied on June 7, 2017 • Show version history

Steve,

Unless I'm mistaken, it sounds like you might be looking at this backward. I don't think you need the Laserfiche URL to "catch" the value, rather you need your other app to build the URL dynamically.

So, in your example your app would have the "start" and "end" of the URL hard coded along with a variable value, and it would concatenate everything to build the final URL.

 

Try something like this:

String input = "311-15-123";

String URL = "https://lfweb.ourdomain.us/laserfiche/Browse.aspx?repo=OurRepository#?view=search&search=%7BLF:Basic%20~%3D%20%22" + input + "%22,%20option%3D%22DFNLT%22%7D%20%26%20%7BLF:Name%3D%22*%22,%20Type%3D%22DF%22%7D%20%26%20%7BLF:LOOKIN%3D%22%5CAssessor%5CPRC%22%7D";

Then you output the dynamically-generated URL from your application.

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replied on June 8, 2017

Thank you Jason, I'll give that a try.

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