How do i get specific value from JSON string

[Goku (IamBatman)]

So am using Api to extract weather details of partiular cities
and i want to extract particular fields such as temp
Below is my JSON

{
“coord”: {
“lon”: -0.1257,
“lat”: 51.5085
},
“weather”: [
{
“id”: 804,
“main”: “Clouds”,
“description”: “overcast clouds”,
“icon”: “04d”
}
],
“base”: “stations”,
“main”: {
“temp”: 20.04,
“feels_like”: 20.08,
“temp_min”: 18.49,
“temp_max”: 21.67,
“pressure”: 1006,
“humidity”: 76
},
“visibility”: 10000,
“wind”: {
“speed”: 4.63,
“deg”: 260
},
“clouds”: {
“all”: 100
},
“dt”: 1690544725,
“sys”: {
“type”: 2,
“id”: 2075535,
“country”: “GB”,
“sunrise”: 1690517856,
“sunset”: 1690574170
},
“timezone”: 3600,
“id”: 2643743,
“name”: “London”,
“cod”: 200
}

How do i do it without using any regex of string manipulations

[Nguyen_Van_Luong1 (Nguyen Van Luong)]

Hi @Goku ,
Use Deserialize JSON activity. It will give you an out put of type JsonObject. Lets us say you store the JsonObject to a variable jsonObject. Now you can get values from the JsonObject
See more at

How to deserialize JSON in UiPath?

Regards,
LNV

[rlgandu (Rajyalakshmi Gandu)]

@Goku

Add an “HTTP Request” activity to call the API and store the JSON response in a variable . (let’s call it jsonResponse ).
Add a “Deserialize JSON” activity after the “HTTP Request” activity.
Create a variable to store the deserialized JSON data (let’s call it weatherData ) and set it as the “Output” property of the “Deserialize JSON” activity.
Assign (temperature = weatherData(“main”)(“temp”).ToString())

[Goku (IamBatman)]

After Deserializing i get the object as below screenshot

How do i extract specific values from this

[rlgandu (Rajyalakshmi Gandu)]

@Goku

UiPath Json Deserialize | UiPath Json |UiPath Json Parsing | UiPath Tutorial

[Anil_G (Anil Gorthi)]

@Goku

1. Deserialize jaon and select type argument as jObject…say output variable is jobj
2. Now use jobj("main")("temp").ToString to get the temperature value

Cheers

[shravan.prajapati]

Please try this approach for your data. Hope it will works.

[Rajasekar_Ramachandran (Rajasekar R)]

You can use the deserialize JSON activity and it will return the JObject
orElse
you can directly parse the string into the JObject in a assign.

Example:
yourJObjectVariable = JObject.Parse(YourApiStringValue)
Then,
yourJObjectVariable(“main”)(“temp”).tostring

[Shubham_Dutta]

Hi @Goku,

try with this expression once after deserializing JSON file
lets take like the output variable of Deserializa JSON activity as out_json
–now in a write line activity mention as
out_json(“main”)(“temp”)(0).tostring

this would give you the value

Regards,
SD

[Aniket_Dandade]

Hello please refer below screeshot for your query:
and if you like my answer please don’t forget to mark it as a solution.