keep the output from a post request inside a string variable - rust

Question

Hi I explain the problem
I'm actually sending a request to the OpenApi and the reply is an url that is basically an image generated by OpenAI.
I would like to keep this url inside a variable after sending a request rather than get this url in the terminal with the println!("{:?}", res);
I need to use this url in an other file after. it's why it's more interesting for me to keep this url in a string variable after the request.
use exitfailure::ExitFailure;
use reqwest::{
header::{ACCEPT, AUTHORIZATION, CONTENT_TYPE},
Url,
};
use serde_derive::{Deserialize, Serialize};
use std::collections::hash_map::*;
// use std::env;
#[derive(Serialize, Deserialize, Debug)]
struct GenerationImage {}
#[derive(Serialize, Deserialize, Debug)]
struct Data {
url: String,
}
#[derive(Serialize, Deserialize, Debug)]
struct Response {
data: [Data; 1],
}
impl GenerationImage {
async fn post(api_key: &str) -> Result<(), ExitFailure> {
let key = format!("Bearer {}", api_key);
let prompt = "a cat behind a tree".to_string();
let url = "https://api.openai.com/v1/images/generations";
let mut map = HashMap::new();
map.insert("prompt", prompt);
let endpoint = Url::parse(url)?;
let client = reqwest::Client::new();
let res = client
.post(endpoint)
.header(AUTHORIZATION, key)
.header(CONTENT_TYPE, "application/json")
.header(ACCEPT, "application/json")
.json(&map)
.send()
.await?
.json::<Response>()
.await;
println!("{:?}", res);
Ok(())
}
}
fn main() {
let api_key = "API";
GenerationImage::post(&api_key);
}

Answer 1

To convert an item that implements Debug into the corresponding String you can do this:
let s = format!("{:?}", res);