Getting Data From Json

Hi,

I need to get Sno,Name,Age data but there is one extra data was added at top.(Attached screenshot for reference)

image

Thanks,
Rishi

1 Like

please go through this video you can parse the json by this way

Hey @Rishik_Chowdary

Did you tried using Deserialize Json activity ?

Thanks
#nK

do start as following
grafik

typeargument from for each activity is JObject

HI @ppr

It was getting error like Value Cant be null…Parameter Name -Value

Yes @Nithinkrishna

we need more context / details, as statement is too isolated.

what have you implemented so far (can you share the json and some screenshots from the implementation)

Had you debugged and checked the variables?

Yes @ppr

i have deserialize and stored in to one variable and i have taken for each in that i have taken assign acivity

For each type argument - newtonsoft.json.linq.jobject

would help us. The incomplete parts from the screenshot are not common with the json snippet from above.

in general the access is not retrievieng the property, thats why it will become null

Kindly note: as better we are served with the relevant information we can help. It will reduce ping pongs and will lead faster to the solution. Thanks for supporting on us on this

Thanks @ppr

Here is the attached json file

JsonFile.txt (686 Bytes)

we can access the value

kindly note: inspections are faster done within the debugging panels instead of using message boxes

But what my question is i dont want that first json values because same headers getting reflected as values

From down i need to get the data and append in to excel

Hope you understand…Thanks

From above

we understood value access

we helped / demonstrated the access

When first JArray item should be ommited, then try frollowing:
For each Activity: item in myJArray.Skip(1)

feel free to redescribe your case / what you are looking for more in detail. Thanks

1 Like

Hey @Rishik_Chowdary

So is that resolved or still you are facing some issues ?

Thanks
#nK

Assign :

SNo = JsonArray(1)(“Sno”).Tostring

Name = JsonArray(1)(“Name”).Tostring

Age = JsonArray(1)(“Age”).Tostring

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