'How to trigger a response when another response done successfuly?
i m getting a link from a response data for another response. how can i wait for finish this first response ?
Response responsePostImage =
await dio.get("responseLink$imgId");
var parsedPostImage = jsonDecode("$responsePostImage");
blogModel.imagePostImage =
parsedPostImage["media_details"]["sizes"]["medium"]["source_url"];
var parsedPpp = parsedPostImage["_links"]["author"];
blogModel.imagePpUrl = parsedPpp[0]["href"];
Response responseGetPp = await dio.get(blogModel.imagePp);
var parsedGetPp = jsonDecode("$responseGetPp");
blogModel.imagePp = parsedGetPp["avatar_urls"]["48"];
print(parsedGetPp["avatar_urls"]["48"]);
Solution 1:[1]
I think you can check the http response success status code (200) then call the next api.
Response responsePostImage = await dio.get("responseLink$imgId");
if(responsePostImage.statusCode == 200) {
var parsedPostImage = jsonDecode("$responsePostImage");
blogModel.imagePostImage = parsedPostImage["media_details"]["sizes"]["medium"]["source_url"];
var parsedPpp = parsedPostImage["_links"]["author"];
blogModel.imagePpUrl = parsedPpp[0]["href"];
Response responseGetPp = await dio.get(blogModel.imagePp);
if(responseGetPp.statusCode == 200) {
var parsedGetPp = jsonDecode("$responseGetPp");
blogModel.imagePp = parsedGetPp["avatar_urls"]["48"];
print(parsedGetPp["avatar_urls"]["48"]);
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Zakaria Hossain |
