Need to Read JSON data and display last name of the employee

[S_S_Kumar]

Read JSON data and display last name of the employee whose ID is “2” in a message box.

5. {
6. “employee”: [

8.   { 
   

9.      "id":”1”, 
   

10.     "name": "Tom", 
    

11.     "lastname": "Prince" 
    

12.  }, 
    

14.  { 
    

15.     "id":"2", 
    

16.     "name": "Nick", 
    

17.     "lastname": "Johnson" 
    

18.  } 
    

19. ]
20. }

Please help me to create a workflow on this from end to end

[Palaniyappan (Palaniyappan P)]

Hi

Welcome to uipath forum

Use the deserialise Json array activity

Deserialize Json

For that go to design tab → manage packages → all packages → search for UiPath.web.Activity and install it

And steps involved are as follows

1. Read the string input and save it in a variable named Strinput

2. Then use a DESERIALISE JSON activity and pass the above string variable as input and type argument will be Newtonsoft.Json.LinqJObject result of this activity will be jsonobject

3. Use a assign activity like this

Jsonstring = jsonobject(“employee”).ToString

4. Then Use Deserialize Json Array activity with Type Argument and pass the Jsonstring in input and say the output of this activity will be Jsonarray

Deserialize Json Array

5. Now use a FOR EACH activity and pass the above variable Jsonarray and change the type argument as Newtonsoft.Json.LinqJObject

6. Inside the loop use a IF condition like this

item(“id”).ToString.Contains(“2”)

If true it will go to then block where use a assign activity like this

str_output = item(“lastname”).ToString

In fact we have got examples on how to use those two deserialise activity in that doc itself
Please refer that for more details

Cheers @S_S_Kumar

[ppr (Peter Preuss)]

@Palaniyappan

5. Now use a FOR EACH activity and pass the above variable Jsonarray and change the type argument as Newtonsoft.Json.LinqJObject
6. Inside the loop use a IF condition like this

item(“id”).ToString.Contains(“2”)

If true it will go to then block where use a assign activity like this

str_output = item(“lastname”).ToString

As an alternate we can filter to the person directly without for each:

Variables:

Flow:

Find starter help here:
ppr_JSONBox_SSKumar.xaml (6.8 KB)