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how to convert json string to json object in dart flutter?

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0

I have string like this,

id:1, name: lorem ipsum, address: dolor set amet

And I need to convert that string to json, how I can do it in dart flutter? thank you so much for your help.

json dart flutter

share|improve this question

asked Mar 22 at 3:49

AshtavAshtav

1517

add a comment | 

0

I have string like this,

id:1, name: lorem ipsum, address: dolor set amet

And I need to convert that string to json, how I can do it in dart flutter? thank you so much for your help.

json dart flutter

share|improve this question

asked Mar 22 at 3:49

AshtavAshtav

1517

add a comment | 

I have string like this,

id:1, name: lorem ipsum, address: dolor set amet

And I need to convert that string to json, how I can do it in dart flutter? thank you so much for your help.

json dart flutter

share|improve this question

asked Mar 22 at 3:49

AshtavAshtav

1517

id:1, name: lorem ipsum, address: dolor set amet

json dart flutter

json dart flutter

share|improve this question

asked Mar 22 at 3:49

AshtavAshtav

1517

share|improve this question

asked Mar 22 at 3:49

AshtavAshtav

1517

asked Mar 22 at 3:49

AshtavAshtav

1517

asked Mar 22 at 3:49

AshtavAshtav

1517

2 Answers
2

active

oldest

votes

0

You have to use json.decode. It takes in a json object and let you handle the nested key value pairs. I'll write you an example

// actual data sent is success: true, data:token:'token'
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) 
 //Token is nested inside data field so it goes one deeper.
 final String token = body['data']['token'];

 return "success": true, "token": token;

share|improve this answer

answered Mar 22 at 4:14

forJforJ

1,2692825

add a comment | 

0

You must import dart:encode libary. Then use the jsonDecode function, that will produce a dynamic similar to a Map

https://api.dartlang.org/stable/2.2.0/dart-convert/dart-convert-library.html

share|improve this answer

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have been try this, Map res = jsonDecode(sharedPreferences.getString('jsonString')); but I got this error: FormatException (FormatException: Unexpected character
  
  – Ashtav
  Mar 22 at 4:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  a Map is distinct of dynamic
  
  – Ernesto Campohermoso
  Mar 24 at 17:16
  

  
  
  
  

add a comment | 

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0

You have to use json.decode. It takes in a json object and let you handle the nested key value pairs. I'll write you an example

// actual data sent is success: true, data:token:'token'
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) 
 //Token is nested inside data field so it goes one deeper.
 final String token = body['data']['token'];

 return "success": true, "token": token;

share|improve this answer

answered Mar 22 at 4:14

forJforJ

1,2692825

add a comment | 

0

You have to use json.decode. It takes in a json object and let you handle the nested key value pairs. I'll write you an example

// actual data sent is success: true, data:token:'token'
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) 
 //Token is nested inside data field so it goes one deeper.
 final String token = body['data']['token'];

 return "success": true, "token": token;

share|improve this answer

answered Mar 22 at 4:14

forJforJ

1,2692825

add a comment | 

You have to use json.decode. It takes in a json object and let you handle the nested key value pairs. I'll write you an example

// actual data sent is success: true, data:token:'token'
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) 
 //Token is nested inside data field so it goes one deeper.
 final String token = body['data']['token'];

 return "success": true, "token": token;

share|improve this answer

answered Mar 22 at 4:14

forJforJ

1,2692825

// actual data sent is success: true, data:token:'token'
final response = await client.post(url, body: reqBody);

// Notice how you have to call body from the response if you are using http to retrieve json
final body = json.decode(response.body);

// This is how you get success value out of the actual json
if (body['success']) 
 //Token is nested inside data field so it goes one deeper.
 final String token = body['data']['token'];

 return "success": true, "token": token;

share|improve this answer

answered Mar 22 at 4:14

forJforJ

1,2692825

answered Mar 22 at 4:14

forJforJ

1,2692825

answered Mar 22 at 4:14

forJforJ

1,2692825

0

You must import dart:encode libary. Then use the jsonDecode function, that will produce a dynamic similar to a Map

https://api.dartlang.org/stable/2.2.0/dart-convert/dart-convert-library.html

share|improve this answer

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have been try this, Map res = jsonDecode(sharedPreferences.getString('jsonString')); but I got this error: FormatException (FormatException: Unexpected character
  
  – Ashtav
  Mar 22 at 4:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  a Map is distinct of dynamic
  
  – Ernesto Campohermoso
  Mar 24 at 17:16
  

  
  
  
  

add a comment | 

0

You must import dart:encode libary. Then use the jsonDecode function, that will produce a dynamic similar to a Map

https://api.dartlang.org/stable/2.2.0/dart-convert/dart-convert-library.html

share|improve this answer

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I have been try this, Map res = jsonDecode(sharedPreferences.getString('jsonString')); but I got this error: FormatException (FormatException: Unexpected character
  
  – Ashtav
  Mar 22 at 4:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  a Map is distinct of dynamic
  
  – Ernesto Campohermoso
  Mar 24 at 17:16
  

  
  
  
  

add a comment | 

You must import dart:encode libary. Then use the jsonDecode function, that will produce a dynamic similar to a Map

https://api.dartlang.org/stable/2.2.0/dart-convert/dart-convert-library.html

share|improve this answer

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

share|improve this answer

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945

answered Mar 22 at 3:55

Ernesto CampohermosoErnesto Campohermoso

6,14512945