Question

现在我正在做这件事：

self.response.headers['Content-Type'] = 'application/json'
self.response.out.write('{"success": "some var", "payload": "some var"}')

使用某个库有没有更好的方法呢？

Answer 1

是的，您应该使用Python 2.7支持的json library：

import json

self.response.headers['Content-Type'] = 'application/json'   
obj = {
  'success': 'some var', 
  'payload': 'some var',
} 
self.response.out.write(json.dumps(obj))

Answer 2

webapp2为json模块提供了一个方便的包装：如果可用，它将使用simplejson，如果可用，则使用Python＆gt; = 2.6的json模块，以及作为最后一个资源的django.utils.simplejson模块App Engine。

http://webapp2.readthedocs.io/en/latest/api/webapp2_extras/json.html

from webapp2_extras import json

self.response.content_type = 'application/json'
obj = {
    'success': 'some var', 
    'payload': 'some var',
  } 
self.response.write(json.encode(obj))

Answer 3

python本身有一个json module，它将确保您的JSON格式正确，手写JSON更容易出错。

import json
self.response.headers['Content-Type'] = 'application/json'   
json.dump({"success":somevar,"payload":someothervar},self.response.out)

Answer 4

我通常这样使用：

import java.util.Scanner;
public class ConvPhoneNumTwo{
  public static void main(String[] args){
    String s = getInput();


    printPhone(convertToDigit(s));
  }   

 public static String getInput() { 
      Scanner scan = new Scanner(System.in);
      String s1 = "";
      boolean length = false;
      while (length==false){ 
        System.out.println(" Please enter a telephone number in this format ###-AAA-AAAA") ;
         s1 = scan.nextLine();
        if(s1.length() != 12 ){
          System.out.println( "this is an invalid choice try again Plz..."); 
           length = false;

        }
        else{ 
          length = true; 
        }
      }
      return s1; 
    }

 public static String convertToDigit(String s010){
    s010 = s010.toLowerCase();
    String s001= s010.substring(0,3);
    String s002 = s010.substring(4,6);
    String s003 = s010.substring(8,12);
    String s1 = (s001+s002+s003);
    String s2 = "";

    // Exceptions to our problem to stop invalid inputs 
    if (s1 == null) {
      System.out.print (" invalid input null thing"); 
    }
    if (s1.length() != 10){
      System.out.print (" invalid input"); 
    }

    String s6 = "";
    for (int i=0; i < s1.length(); i++) {


      if (Character.isDigit(s1.charAt(i))){
        s2 += s1.charAt(i); 
      }

      //sorting of the letter inputs 
      char ch = s1.charAt(i); 

      if (ch == 'a'||ch=='b'||ch== 'c'){
        s2 += "2"; 
      }
      if (ch == 'd'||ch=='e'||ch=='f'){
        s2 += "3"; 
      }
      if (ch == 'g'||ch=='h'||ch=='i'){
        s2 += "4"; 
      }
      if (ch == 'j'||ch=='k'||ch=='l'){
        s2 += "5"; 
      }
      if (ch == 'm'||ch=='n'||ch=='o'){
        s2 += "6"; 
      }
      if (ch == 'p'||ch=='q'||ch=='r'|| ch=='s'){
        s2 += "7"; 
      }
      if (ch == 't'||ch=='u'||ch=='v'){
        s2 += "8"; 
      }
      if (ch == 'w'||ch=='x'||ch=='y'|| ch=='z')
      {
        s2 += "9"; 
      }
      else{ 
      }

     String s3 = s2.substring(0,3);
    String s4 = s2.substring(3,6);
    String s5 = s2.substring(6);
     s6 = ( s3 +"-"+ s4 + "-"+ s5);
    }

    return s6; 
    }

    public static void printPhone(String message) { //print out whatever string it is given, basically System.out.println(), but through a method instead
    System.out.println(message);
  }
}

Answer 5

import json
import webapp2

def jsonify(**kwargs):
    response = webapp2.Response(content_type="application/json")
    json.dump(kwargs, response.out)
    return response

你想要回复json回复的每个地方......

return jsonify(arg1='val1', arg2='val2')

或

return jsonify({ 'arg1': 'val1', 'arg2': 'val2' })

如何使用app引擎Python webapp2正确输出JSON？

5 个答案: